d
The two opposite cells $A$ and $B$ will cancel two more cells, so net emf will be $n-4 .$ So current is $I=\frac{(n-4) \varepsilon}{n r}$
Now $pd$ across $A$ or $B$ is
$I \varepsilon+I r$ (as they will be in charging state), brgt
$=\varepsilon=\frac{(n-4) \varepsilon}{n}=2 \varepsilon\left(1-\frac{2}{n}\right)=\frac{2 \varepsilon(n-2)}{n}$