Q + Br2/KOH → R is Hofmann bromamide degradation: Amide → Amine (with one less carbon).
R = C6H7N = C6H5NH2 = Aniline
For Q to give aniline, Q = Benzamide (C6H5CONH2)
C6H5CONH2Br2/KOHC6H5NH2
P + hot NH3 → Q: Carboxylic acid + NH3 → Amide
P = Benzoic acid (C6H5COOH)
C6H5COOHNH3,ΔC6H5CONH2
P = Benzoic acid, Q = Benzamide, R = Aniline