Correct answer is (A)

As $P=\frac{1}{3}\frac{mN}{V}{v}_{rms}^2$   ...(i)

where m is the mass of each molecule, N is the total number of molecules, V is the volume of the gas. When mass of all the molecules is halved and their speed is doubled, then the pressure will be

$P^{\prime }=\frac{1}{3}\left(\frac{m}{2}\right)\times \frac{N}{V}\times {\left(2v_{rms}\right)}^2$

$=\frac{2}{3}\frac{mN}{V}{v}_{rms}^2=2P$   (Using (i))