Initially, when both bulbs are at 273 K. the mole in each bulb,
$ = \large\frac{Total \;mole}{2} = \large\frac{P.2V}{2RT} =\large\frac{PV}{RT} = \large\frac{76\times V}{R\times273}$
On heating second bulb some mole of gas are transferred to bulb One till the pressure in two bulbs becomes same.
For bulb 1 $n_1 = \large\frac{P_1 \times V}{R\times273}$
For bulb 2 $n_2 = \large\frac{P_2\times V}{R\times 335}$
Since $n = n_1 + n_2 $
$\therefore \large\frac{76\times V}{R\times273}\times2 = \large\frac{P_1\times V}{R\times273}+\large\frac{P_2\times V}{R\times335}$
also $P_1 = P_2$
$\therefore P_1 = 83.75$ cm Hg
Hence answer is (b)