In case $(i)$, galvanometer shows zero deflection.

$\therefore V_B=V_D$

$\Rightarrow \frac{P}{S}=\frac{Q}{R} \quad \dots(i)$

When battery $B$ and galvanometer $G$ are interchanged, position of galvanometer is as shown below,

Now, ratio of resistances across galvanometer is

$\frac{S}{P} \text { and } \frac{R}{Q}$

As from Eq. $(i)$, $\frac{S}{P}=\frac{R}{Q}$

Hence, galvanometer still shows zero deflection because Wheatstone's bridge is balanced.