For the reaction,
${\mathrm{PCl}}_5\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{PCl}}_3\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$
The forward reaction at constant temperature is favoured by:

$$\begin{gathered} \underset{\mathrm{Mole} \\ \mathrm{At}. \mathrm{eq}. (\mathrm{a}-\mathrm{x})}{{\mathrm{PCl}}_5\left(\mathrm{g}\right)}\rightleftharpoons \underset{ \\ \mathrm{x}}{{\mathrm{PCl}}_3\left(\mathrm{g}\right)}+\underset{ \\ \mathrm{x}}{{\mathrm{Cl}}_2\left(\mathrm{g}\right)} \end{gathered}$$
Now, ${\mathrm{K}}_{\mathrm{c}}=\frac{{\mathrm{x}}^2}{\mathrm{V}(\mathrm{a}-\mathrm{x})}$
Addition of inert gas at constant V has no effect on reactions having $\mathrm{\Delta n}=0\text{ or }\mathrm{\Delta n}\ne 0$. But addition of inert gas at constant P has effect on reactions having $\mathrm{\Delta n}\ne 0$ and no effect, if $\mathrm{\Delta n}=0$. The given reaction has $\mathrm{\Delta n}\ne 0$ and thus, only choice (c) is correct. Also, the effect may be drawn as: On addition of inert gas at constant P, volume increases. To have K_c constant, X must increase.