Compound A, $\mathrm{C}{}_7\mathrm{H}{}_9\mathrm{N}$, gives a positive carbylamine test. On treatment with $\mathrm{NaNO}{}_2/\mathrm{HCl}$ at 273-278 K, then $\mathrm{CuCN}/\mathrm{KCN}$, then hydrolysis with aqueous acid, and finally vigorous oxidation with hot alkaline $\mathrm{KMnO}{}_4$, it gives benzene-1,4-dicarboxylic acid (terephthalic acid). A is:

The carbylamine test shows a primary amine. Diazotisation, $\mathrm{CuCN}$ and hydrolysis convert the amino group into a carboxylic acid group; oxidation of the remaining side chain methyl group to carboxyl gives terephthalic acid only if the two groups are para. So A is 4-methylaniline. Benzylamine, a primary aliphatic amine, would not give a diazonium salt, and N-methylaniline is secondary.