Two geometric isomers of acetophenone oxime are available. The Z isomer (hydroxyl group on the same side as the phenyl group) is treated with PCl5…

In the Beckmann rearrangement the group anti to the leaving group on nitrogen (the activated $\mathrm{N{-}OPCl_4}$ group) migrates from carbon to nitrogen. In the $Z$ isomer the hydroxyl group is syn to phenyl, so it is anti to the methyl group. The methyl group therefore migrates, and after loss of the leaving group and capture by water the product is the amide $\mathrm{PhC(=O)NHCH_3}$ (N-methylbenzamide), not acetanilide. Acid hydrolysis then gives benzoic acid and methylammonium ion. Diazotisation works usefully only for aromatic primary amines: aliphatic diazonium ions decompose at once to alcohols or alkenes with loss of $\mathrm{N_2}$, so no diazonium salt survives for coupling and no azo dye forms with $\beta$-naphthol. The $E$ isomer would instead give acetanilide, then aniline, a stable benzenediazonium salt and a scarlet azo dye. Hence option (A).