In the network shown in figure, the equivalent capacitance between points $P$ and $Q$ is

Simplify from right to left.

Rightmost section: $3\mu F$, $3\mu F$, $3\mu F$ in series (path $B \to C \to D \to E$):
$\dfrac{1}{C_{eq1}} = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} = 1 \Rightarrow C_{eq1} = 1\mu F$

This $1\mu F$ is in parallel with $2\mu F$ (between $B$ and $E$):
$C_{BE} = 1 + 2 = 3\mu F$

Next section: $3\mu F$, $3\mu F$ ($C_{BE}$), $3\mu F$ in series (path $A \to B \to E \to F$):
$\dfrac{1}{C_{eq2}} = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} = 1 \Rightarrow C_{eq2} = 1\mu F$

This $1\mu F$ is in parallel with $2\mu F$ (between $A$ and $F$):
$C_{AF} = 1 + 2 = 3\mu F$

Final section: $3\mu F$ (at $P$), $3\mu F$ ($C_{AF}$), $3\mu F$ (at $Q$) in series:
$\dfrac{1}{C_{PQ}} = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} = 1 \Rightarrow C_{PQ} = 1\mu F$

Correct option: (1)