$\left[\mathrm{P}\right]$ on treatment with ${\mathrm{Br}}_2/{\mathrm{FeBr}}_3$ in ${\mathrm{CCl}}_4$ produced a single isomer ${\mathrm{C}}_8{\mathrm{H}}_7{\mathrm{O}}_2\mathrm{Br}$ wile heating $\left[\mathrm{P}\right]$ with sodalime gives toluene. The compound $\left[\mathrm{P}\right]$ is:

Isomer ${\mathrm{C}}_8{\mathrm{H}}_7{\mathrm{O}}_2\mathrm{Br}$ on heating with sodalime gives toluene,hence it is an acid.Moreover since it has Degree of unsaturation 5 therefore it is an aromatic carboxylic acid.Since Carboxylic acid on heating gives toluene ,therefore it is having $-{\mathrm{CH}}_3$ group attached to the benzene ring.

Now it forms single isomer on halogenation, therefore $-{\mathrm{CH}}_3$ group must be attached at para position with respect to $-\mathrm{COOH}$ group.

Hence correct option is 1.