Benzamide is treated with Br2/NaOH (aqueous, warm) to give A. Compound A reacts with NaNO2/HCl at 273–278 K to give B, which on warming with water…

Hofmann bromamide degradation: $\mathrm{C_6H_5CONH_2\xrightarrow{Br_2/NaOH}C_6H_5NH_2}$ ($\mathbf{A}$, aniline; one carbon is lost as carbonate). Diazotisation gives benzenediazonium chloride ($\mathbf{B}$); warming with water replaces $\mathrm{N_2^+}$ by $\mathrm{OH}$, evolving $\mathrm{N_2}$ and giving phenol ($\mathbf{C}$). Reimer-Tiemann reaction of phenol: $\mathrm{CHCl_3/NaOH}$ generates dichlorocarbene, which attacks the phenoxide ion mainly at the ortho position, so the major product is 2-hydroxybenzaldehyde (salicylaldehyde), with 4-hydroxybenzaldehyde minor. Salicylaldehyde has an intramolecular O-H...O=C hydrogen bond, so molecules do not associate with one another and it is more volatile and steam volatile; the 4-isomer is intermolecularly associated. The carbylamine reaction needs a primary amine, but $\mathbf{C}$ is a phenol. Hence option (A) is correct, (B) and (C) misplace the substituent or reverse the volatility argument.