VPQ in the figure.

C = $\frac{2 \times 4}{2+4}=\frac{4}{3}$ µF in one loop.

q = $12 \times \frac{4}{3}$ = 16 µC across one set of 2 µF & 4 µF capacitors.

∴ $V_P=12-\frac{16}{4} = 8 V$

$V_Q=12-\frac{16}{2}=4 V$

∴ V_P – V_Q = 8 – 4 = 4 V

∴ (C)