Answer
(a) decrease
(b) III
Work Step by Step
(a) We know that
$\frac{1}{C_{12}}=\frac{1}{C_1}+\frac{1}{C_2}$
$\implies C_{12}=\frac{1}{C}+\frac{1}{C}$
$C_{12}=\frac{C}{2}$
Thus, if another capacitor is connected then the equivalent capacitance is $C_{effective}=\frac{C}{3}$ and thus the equivalent capacitance is decreased.
(b) We know that the best explanation is option (III) -- that is, adding a capacitor in series decreases the equivalent capacitance since each capacitor now has less voltage across it, and hence stores less charge.
