Four capacitors are connected in the circuit as shown.

The equivalent capacitance between P and Q will be:

8 μF

The correct answer is Option (1) → 8 μF

Let node $A=P$ (left & bottom rail), node $B$ (top rail), node $C=Q$ (right rail).

Between $A$ and $B$: two $6\,\mu\text{F}$ in parallel $\Rightarrow C_{AB}=6+6=12\,\mu\text{F}$.

Between $B$ and $C$: $C_{BC}=6\,\mu\text{F}$.

Between $A$ and $C$: $C_{AC}=4\,\mu\text{F}$.

Path $A\!\to\!B\!\to\!C$ is series: $C_s=\frac{C_{AB}\,C_{BC}}{C_{AB}+C_{BC}}=\frac{12\times 6}{12+6}=4\,\mu\text{F}$.

Two paths between $A$ and $C$ are in parallel: $C_{\text{eq}}=C_{AC}+C_s=4+4=8\,\mu\text{F}$.