At balance, galvanometer current $I_g = 0$.

Applying KCL:
$I_1 = I_3$ (through $P$ and $R$), $I_2 = I_4$ (through $Q$ and $S$)

Applying KVL to loop ABDA:
$$I_1 P = I_2 Q \quad \cdots (1)$$

Applying KVL to loop BCDB:
$$I_3 R = I_4 S \quad \cdots (2)$$

Dividing (1) by (2) and using $I_1 = I_3$, $I_2 = I_4$:
$$\frac{P}{Q} = \frac{R}{S}$$
This is the balance condition of the Wheatstone bridge.