Why does reducing the pressure in an equilibrium reaction favour the side with more moles of gas?
For your reaction, $\Delta n_g = 3-(1+1)=1$ (assuming all A,B,C are gases). $K_p = \frac{[C]^3}{[A][B]} (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$, where $\Sigma n$ is the total number of moles of gases in the reaction mixture(even of those species which are not taking part in the reaction). Here, it is a constant.

When pressure is decreased($P_{total}$), $ (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$ decreases as $\Delta n_g &gt;0$.

In order to keep $K_p$ constant (it is always constant at a given temperature), $[C]$ has to increase or $[A],[B]$ has to decrease. Therefore the reaction proceeds/shifts more in the forward direction, which has more number of gaseous moles.

Similarly, increase in pressure favours the side with lesser number of gaseous moles.