The Physical state of a gas is represented by P, V and T in Vessel A and 2P,$\frac{\mathrm{V}}{4}$2T in vessel B. The ratio of number of molecules in A and B is

$$\begin{gathered} {\mathrm{P}}_1=\mathrm{P}; {\mathrm{P}}_2=2\mathrm{P}; \\ {\mathrm{V}}_1=\mathrm{V}; {\mathrm{V}}_2=\mathrm{V}/4 \\ {\mathrm{T}}_1=\mathrm{T}; {\mathrm{T}}_2=2\mathrm{T} \end{gathered}$$
We Know that,
$$\begin{gathered} PV=\mu RT \\ {P}_1{V}_1={\mu }_{A}R{T}_1..........\left(1\right) \\ {P}_2{V}_2={\mu }_{B}R{T}_2...........\left(2\right) \end{gathered}$$
Taking ratio of Eq. (1) and Eq (2), we get,
$$\begin{gathered} \frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{P}}_2{\mathrm{V}}_2}=\frac{{\mathrm{\mu }}_{\mathrm{A}}{\mathrm{RT}}_1}{{\mathrm{\mu }}_{\mathrm{B}}{\mathrm{RT}}_2} \\ \frac{PV}{2P(V/4)}=\frac{{\mathrm{\mu }}_{\mathrm{A}}{\mathrm{RT}}_{ }}{{\mathrm{\mu }}_{\mathrm{B}}\mathrm{R}\left(2\mathrm{T}\right)} \\ 2=\frac{{\mathrm{\mu }}_{\mathrm{A}}}{2{\mathrm{\mu }}_{\mathrm{B}}} \\ \frac{{\mathrm{\mu }}_{\mathrm{A}}}{{\mathrm{\mu }}_{\mathrm{B}}}=\frac{4}{1}\Rightarrow {\mathrm{\mu }}_{\mathrm{A}}={\mathrm{\mu }}_{\mathrm{B}}=4:1 \end{gathered}$$