Yes, but the tidal torques will prevent the planet from being locked either to the star or moon.

Suppose the orbital paths are both prograde and coplanar and the planet obliquity is zero. Then tidal torque exerted on the planet by its moon should be:

$$N_m=\frac{9}{4}k\frac{Gm_m^2A^5}{r_m^6}\sin{(2\alpha_m)}$$

Here, $m_m$ is the moon mass, $G$ is the standard gravitational parameter, $A$ is the planetary radius, $r_m$ is the distance between the moon and planet, $\alpha_m$ is the tidal lag angle, and $k$ is a factor related to the Love numbers and rigidity of the planet. Note that if the planet is already tidally locked to the moon, then $\alpha_m = 0$, so there is no torque being exerted on the planet.

Similarly, the tidal torque exerted on the planet by the star is: $$N_s=\frac{9}{4}k\frac{Gm_s^2A^5}{r_s^6}\sin{(2\alpha_s)}$$

Here, $m_s$ is the star mass, $r_s$ is the distance between the star and planet, and $\alpha_s$ is the tidal lag angle.

The moon's orbital period around the planet will be much smaller than the planet's orbital period around the star. So $\alpha_s \ne 0$. This means there will be significant torque on the planet in the opposite direction of its rotation, due to the star. As this torque slows the rotational speed of the planet, the planet will move out of tidal lock with the moon. This will increase the absolute value of $\alpha_m$, and the moon will begin exerting tidal torque on the planet in the direction of its rotation.

Then $N_m$ and $N_s$ will be rotation torques in opposition. For some rotation rate for the planet, they will be equal and opposite, and the planet's rotation rate will stabilize, without it being tidally locked to either its moon or star. Venus is in a similar situation with the Sun. The tidal torque on Venus due to the Sun is in opposition to the torque on Venus due to atmospheric thermal tides, which prevents it from being tidally locked.

Notes:

1. Tidal torques conserve angular momentum, so rotational momentum is transferred into orbital momentum, causing planets' and moons' orbits to migrate. If a moon loses too much orbital momentum it will crash into the planet. If it gains too much, it will exit the planet's hill sphere and be ejected (or crash back into another body). See Alavarado-Montes and Sucerquia 2019. For close-in planets, tidal torque is much higher than for far-out planets, scaling as $1/r^6$. This could be why we have observed no exo-moons on the abundant close-in, hot-Jovian exoplanets.

2. The above answer only considers planetary torques cause by the moon's gravity on the moon generated tidal bulge, and the star's gravity on the star generated tidal bulge. It doesn't take into account the planetary torques due to each body on the other body's tidal bulge. That is because these cross-bodied torques will pass through every lag angle each time the planet orbits the star. However, certain resonances in lunar and stellar rotation rates could change quickly change orbital parameters and destabilize the 3-body system.

Yes, but the tidal torques will prevent the planet from being locked either to the star or moon.

Suppose the orbital paths are both prograde and coplanar and the planet obliquity is zero. Then tidal torque exerted on the planet by its moon should be:

$$N_m=\frac{9}{4}k\frac{Gm_m^2A^5}{r_m^6}\sin{(2\alpha_m)}$$

Here, $m_m$ is the moon mass, $G$ is the standard gravitational parameter, $A$ is the planetary radius, $r_m$ is the distance between the moon and planet, $\alpha_m$ is the tidal lag angle, and $k$ is the quadrupole tidal Love number of the planet. Note that if the planet is already tidally locked to the moon, then $\alpha_m = 0$, so there is no torque being exerted on the planet.

Similarly, the tidal torque exerted on the planet by the star is: $$N_s=\frac{9}{4}k\frac{Gm_s^2A^5}{r_s^6}\sin{(2\alpha_s)}$$

Here, $m_s$ is the star mass, $r_s$ is the distance between the star and planet, and $\alpha_s$ is the tidal lag angle.

The moon's orbital period around the planet will be much smaller than the planet's orbital period around the star. So $\alpha_s \ne 0$. This means there will be significant torque on the planet in the opposite direction of its rotation, due to the star. As this torque slows the rotational speed of the planet, the planet will move out of tidal lock with the moon. This will increase the absolute value of $\alpha_m$, and the moon will begin exerting tidal torque on the planet in the direction of its rotation.

Then $N_m$ and $N_s$ will be rotation torques in opposition. For some rotation rate for the planet, they will be equal and opposite, and the planet's rotation rate will stabilize, without it being tidally locked to either its moon or star. Venus is in a similar situation with the Sun. The tidal torque on Venus due to the Sun is in opposition to the torque on Venus due to atmospheric thermal tides, which prevents it from being tidally locked.

Notes:

1. Tidal torques conserve angular momentum, so rotational momentum is transferred into orbital momentum, causing planets' and moons' orbits to migrate. If a moon loses too much orbital momentum it will crash into the planet. If it gains too much, it will exit the planet's hill sphere and be ejected (or crash back into another body). See Alavarado-Montes and Sucerquia 2019. For close-in planets, tidal torque is much higher than for far-out planets, scaling as $1/r^6$. This could be why we have observed no exo-moons on the abundant close-in, hot-Jovian exoplanets.

2. The above answer only considers planetary torques cause by the moon's gravity on the moon generated tidal bulge, and the star's gravity on the star generated tidal bulge. It doesn't take into account the planetary torques due to each body on the other body's tidal bulge. That is because these cross-bodied torques will pass through every lag angle each time the planet orbits the star. However, certain resonances in lunar and stellar rotation rates could change quickly change orbital parameters and destabilize the 3-body system.