What is wrong with my calculations of Jupiter's apparent magnitude? I was searching for a formula to calculate the apparent magnitude of a planet knowing its physical characteristics and stumbled upon this answer.
The final formula given is:
$$m_{planet} = V_{planet} + 5 \log_{10}\left( d_{e-p} \right) - 5$$
In which,
$$ V_{planet}=-2.5 \log_{10}\left(a_p \frac{ r_p^2 }{ 4 d_s^2 } \right) - V_{sun}$$
where:

$m_{planet}$ is the planet in question's apparent magnitude,

$V_{planet}$ is the planet's absolute magnitude,

$a_{p}$ is the bond albedo of the planet,

$r_{p}$ is the radius of the planet,

$d_{s}$ is the distance of the planet from its star,

$d_{e-p}$ is the distance between the planet and the observer (here on Earth) in parsecs, and,

$V_{sun}$ is the star's absolute magnitude.



To test it out I tried to calculate the apparent magnitude of Jupiter as seen from Earth at an average opposition. The values then are:

$a_{p} = 0.343$,

$r_{p} = 69,911,000$ m,

$d_{s} = 778.57 \times 10^9$ m (5.204 AU),

$d_{e-p} = 628.97 \times 10^9$ m $= 0.0000203835294968$ parsecs (4.204 AU)

$V_{sun} = 4.83$


All the above values were taken from NASA's Jupiter fact sheet.
Plugging in the numbers and I get $m_{planet} = -10.38$, which is much brighter than the actual apparent magnitude of Jupiter at average opposition of $-2.x$. Where did I go wrong?
I think that should be $+ V_{Sun}$ in your second equation, not $- V_{Sun}$.  Otherwise, a fainter Sun (larger magnitude) would give you a brighter Jupiter (smaller magnitude).   Those need to go in the same direction.
Applying that change (assuming the rest of the calculation is right) would increase your answer by $2 \times 4.83 = 9.66$, which gets you much closer to the right magnitude.
Your estimate will now be a bit on the faint side - not sure why - but at least a lot closer than before.