Lets assume we're dealing with a superior planet, which is to say, the planet is orbiting at a greater distance than the Earth. This effectively ensures that there is no planetary phase to deal with.
Now, the sun has a luminosity of $L_{sun}$ such that the Solar flux as seen by the planet
is
$$
f_{planet} = \frac{ L_{sun} }{ 4 \pi d_s^2 }.
$$
The cross section of the planet is about $A = \pi r_p^2$, so for a given albedo $a_p$ the reflective luminosity of the planet will be
\begin{aligned}
L_{planet} &= a_p f_{planet} A \\
&= a_p \pi r_p^2 \frac{ L_{sun} }{ 4 \pi d_s^2 }.
\end{aligned}
Given that we know the absolute magnitude of the sun, the absolute magnitude of the planet follows as
\begin{aligned}
V_{planet} &= -2.5 \log_{10}\left[ \frac{ L_{planet} }{ L_{sun} } \right] - V_{sun} \\
&= -2.5 \log_{10}\left[ a_p \frac{ r_p^2 }{ 4 d_s^2 } \right] - V_{sun}
\end{aligned}
The apparent magnitude of the planet as seen from Earth can then be calculated as
$$
m_{planet} = V_{planet} + 5 \log_{10}\left[ d_{e-p} \right] - 5
$$
where the distance between Earth and the planet, $d_{e-p}$, should be in parsec and of course depends on the orbital phase of the two respective planets.
So the one additional parameter required was the Solar luminosity. There are probably a bunch of subtle effects with he albedo and the geometry of the system that are not taken into account here, but it should be a fair approximation.