An organic compound (P) on treatment with aqueous ammonia under hot...

Q + $Br_2$/KOH → R is Hofmann bromamide degradation: Amide → Amine (with one less carbon).
R = $C_6H_7N=C_6H_5NH_2$ = Aniline
For Q to give aniline, Q = Benzamide ($C_6H_5CONH_2$)
$C_6H_5CONH_2 \xrightarrow{Br_2/KOH} C_6H_5NH_2$
P + hot $NH_3$ → Q: Carboxylic acid + $NH_3$ → Amide
P = Benzoic acid ($C_6H_5COOH$)
$C_6H_5COOH \xrightarrow{NH_3, \Delta} C_6H_5CONH_2$
P = Benzoic acid, Q = Benzamide, R = Aniline