If ${\mathrm{x}}_1$ and ${\mathrm{x}}_2$ represent the mole fraction of a component $\mathrm{A}$ in the vapour phase and liquid mixture respectively and ${\mathrm{p}}_{\mathrm{A}}^0$ and ${\mathrm{p}}_{\mathrm{B}}^0$ represent vapour pressure of pure $\mathrm{A}$ and pure $\mathrm{B}$ then total vapour pressure of the liquid mixture is

We know that,

In the liquid phase,

Mole fraction of a component $\mathrm{A}$=${\mathrm{x}}_2$

Now, By Raoult's Law

${\mathrm{P}}_{\mathrm{A}}={\mathrm{p}}_{\mathrm{A}}^0{\mathrm{x}}_2$

In the Vapour phase,

Vapour pressure of pure $\mathrm{A}$= ${\mathrm{p}}_{\mathrm{A}}^0$

Mole fraction of a component $\mathrm{A}$ =  ${\mathrm{x}}_1$

Mole fraction of A in Vapour phase$=\frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{total}}}$

${\mathrm{x}}_1=\frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{t}}}=\frac{{\mathrm{p}}_{\mathrm{A}}^0{\mathrm{x}}_2}{{\mathrm{P}}_{\mathrm{t}}}$

Therefore,

${\mathrm{P}}_{\mathrm{t}}=\frac{{\mathrm{p}}_{\mathrm{A}}^0{\mathrm{x}}_2}{{\mathrm{x}}_1}$

Hence, option(b) is the correct answer.