Choose the correct circuit which can achieve the bridge balance.

In the circuit where the diode shorts the network, no balance is possible. In the circuits where the diode D is reverse biased, the arm containing it is open, so the bridge cannot balance. In the circuit where the diode is forward biased (acting as a closed switch with its resistance $R_d$), the bridge can be balanced if $\frac{P}{Q} = \frac{R}{S}$: $\frac{10}{15} = \frac{10}{5 + R_d} \Rightarrow R_d = 10\ \Omega$ So the correct circuit is the one in which the diode is forward biased.