Chemical Equilibrium - Le Chatelier's Principle, Change in Volume
Expanding on what Buck Thorn said in the comments and providing a more rigorous answer.

The partial pressure would change as follows:
$$\begin{array}{c|c|c|c}
&amp; \ce{SO2} &amp; \ce{O2} &amp; \ce{SO3} \\
\hline
\text{Initial state} &amp; P_A &amp; P_B &amp; P_C \\
\text{On halving volume} &amp; 2P_A &amp; 2P_B &amp; 2P_c\\
\text{Final state} &amp; 2P_A-2x &amp; 2P_B-x &amp; 2P_c + 2x\\
\end{array}$$

where $P_A + P_B + P_C = P_1$

Since initial and final states are in equilibrium,
$$ K_p = \frac{P_C^2}{P_A^2 \cdot P_B} = \frac{(2P_C+2x)^2}{(2P_A-2x)^2 \cdot (P_B+x)}$$

We do not need to solve this equation: looking at both the sides, we can see that the value of x would be positive (for a negative x, the numerator would decrease and denominator would comparatively increase due to the $(2P_A - 2x)^2$ term increasing more than the $(P_B + x)$ term due to the square). You could go ahead and solve the equation for a more accurate answer but looking at the options, we can make do with an approximation.

Adding the partial pressures at the final state, we have
$$P_2 = 2(P_A + P_B + P_C) - x = 2P_1 - x &lt; 2P_1$$

Which corresponds to option (d)