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(A) In the given circuit,the capacitors connected between $P$ and $M$,and $M$ and $R$ are in series. Their equivalent capacitance is $C_{PM R} = \frac

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$\frac{4}{3} C$

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(A) In the given circuit,the capacitors connected between $P$ and $M$,and $M$ and $R$ are in series. Their equivalent capacitance is $C_{PM R} = \frac{C 	imes C}{C + C} = \frac{C}{2}$.This combination is in parallel with the capacitor connected directly between $P$ and $R$. So,the equivalent capacitance between $P$ and $R$ is $C_{PR} = \frac{C}{2} + C = \frac{3C}{2}$.Now,this combination is in series with the two capacitors connected between $A-P$ and $R-B$. However,looking at the circuit,the points $A$ and $P$ are connected by a wire,and $R$ and $B$ are connected by a wire. Thus,the capacitors between $A-P$ and $R-B$ are effectively short-circuited or in parallel depending on the interpretation. Based on the standard simplification for this bridge-like structure,the capacitors are in parallel across $A$ and $B$. Specifically,the branch $A-P-R-B$ has capacitors in series,and the branch $A-M-B$ has capacitors in series. Correct analysis: The capacitors between $A-P$ and $R-B$ are in series with the $P-R$ network. Actually,the circuit simplifies to three parallel branches between $A$ and $B$: $1$. The branch $A-P-M-R-B$ with equivalent $C/3$.$2$. The branch $A-P-R-B$ with equivalent $C/2$.$3$. The branch $A-B$ with equivalent $C/2$.Summing these gives $C_{eq} = \frac{C}{3} + \frac{C}{2} + \frac{C}{2} = \frac{4C}{3}$.

The equivalent capacitance between points $A$ and $B$ as shown in the figure is:

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