b
Here $Q=0$ and $W=0 .$ Therefore from first law of thermodynamics $\Delta U=Q+W=0$
$\therefore$ Internal energy of the system with partition $=$ Internal energy of the system without partition.
$n_{1} C_{v} T_{1}+n_{2} C_{v} T_{2}=\left(n_{1}+n_{2}\right) C_{v} T$
$\therefore T=\frac{n_{1} T_{1}+n_{2} T_{2}}{n_{1}+n_{2}}$
But $n_{1}=\frac{P_{1} V_{1}}{R T_{1}}$ and $n_{2}=\frac{P_{2} V_{2}}{R T_{2}}$
$\therefore T=\frac{\frac{P_{1} V_{1}}{R T_{1}} \times T_{1}+\frac{P_{2} V_{2}}{R T_{2}} \times T_{2}}{\frac{P_{1} V_{1}}{R T_{1}}+\frac{P_{2} V_{2}}{R T_{2}}}=\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}$