$\mathrm{X}$ is a non- volatile solute and $\mathrm{Y}$is a volatile solvent. The following vapour pressures are observed by dissolving $\mathrm{X}$ in $\mathrm{Y}$

$\begin{array}{ll}\mathrm{X}/{\mathrm{mollit}}^{-1}& \mathrm{Y}/\mathrm{mm}\text{ of }\mathrm{Hg}\\ 0.10& {\mathrm{P}}_1\\ 0.25& {\mathrm{P}}_2\\ 0.01& {\mathrm{P}}_3\end{array}$

We know that volatile solvent has high vapour pressure, adding solvent into it results in the decrease of vapour pressure as solute particles added to the solvent increases the boiling point which results decrease in vapour pressure.

Therefore, it means higher the concentration lower the vapour pressure.

i.e., concentration is inversely proportional to vapour pressure.

According to question,

From a given concentration 0.25 is the highest concentration which means it has the lowest vapour pressure.

0.01 is the lowest concentration which means it has the highest vapour pressure.

0.1 is midst concentration which means it has midst vapour pressure.

Hence, ${\mathrm{P}}_2<{\mathrm{P}}_1<{\mathrm{P}}_3$

Therefore, option(d) is correct.