At same temperature ${\mathrm{N}}_2{\mathrm{O}}_4$is dissociated to 40% and 50% at total pressure ${\mathrm{P}}_1$and ${\mathrm{P}}_2$ atm respectively in ${\mathrm{NO}}_2$. 

 

As temperature is constant,

$\begin{array}{l}{K}_{P_1}=K_{P_2}\\ \frac{{\alpha }_1^2}{1-{\alpha }_1^2}\times P_1=\frac{{\alpha }_2^2}{1-{\alpha }_2^2}\times P_2\\ \frac{P_1}{P_2}=\frac{0.25}{0.75}\times \frac{0.84}{0.16}=\frac{7}{4}\end{array}$

Then the ratio of ${\mathrm{P}}_1$ and ${\mathrm{P}}_2$ is: 7 : 4.

Hence, the correct option is b.