An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with $$Br_{2}$$ and KOH forms compound (R) having molecular formula $$C_{6}H_{7}N$$ Names of P, Q and R respectively are.

Organic compound P reacts with aqueous ammonia (hot) to form Q, which on Hofmann degradation (Br₂/KOH) gives R with molecular formula $$C_6H_7N$$. We need to identify P, Q, and R.

$$C_6H_7N$$ corresponds to aniline ($$C_6H_5NH_2$$): molecular formula check: 6C + 7H + 1N = $$C_6H_7N$$ $$\checkmark$$.

In Hofmann degradation, an amide ($$RCONH_2$$) is converted to an amine ($$RNH_2$$) with one fewer carbon atom:

$$ RCONH_2 + Br_2 + 4KOH \rightarrow RNH_2 + K_2CO_3 + 2KBr + 2H_2O $$

Since R is aniline ($$C_6H_5NH_2$$), Q must be benzamide ($$C_6H_5CONH_2$$), which has one more carbon than aniline.

Q (benzamide) is formed from P by reaction with aqueous ammonia under hot conditions. The reaction of a carboxylic acid with ammonia under heating produces an amide:

$$ C_6H_5COOH + NH_3 \rightarrow C_6H_5COONH_4 \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O $$

Therefore, P is benzoic acid ($$C_6H_5COOH$$).

The correct answer is Option (4): Benzoic acid, benzamide, aniline.