b
Let initial resistance of the wires are $R_{1}$ and $R_{2}$ respectively. Then
$R_{1}^{\prime}+R_{2}^{\prime}=R_{1}+R_{2}$
$\Rightarrow R_{1}\left(1+\alpha_{1} \Delta T\right)+R_{2}\left(1+\alpha_{2} \Delta T\right)$
$=R_{1}+R_{2}$
$\Rightarrow R_{1} \alpha_{1}+R_{2} \alpha_{2}=0$
$\frac{\rho_{1} L_{1}}{A}+\frac{\rho_{2} L_{2}}{A} \alpha_{2}=0$
$\Rightarrow \rho_{1} L_{1} \alpha_{1}+\rho_{2} L_{2} \alpha_{2}=0$