Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are :

\( A = \) \( B = \) \( C = \)

The correct option is B B and C only
Compounds which have higher acidic strength than conjugate acid of sodium bicarbonate (i.e., carbonic acid) will react with $NaHC{O}_3$ to produce $C{O}_2$.
Phenol is a weak acid than carbonic acid so it cannot decompose $NaHC{O}_3$.
Generally, all those acids which are more acidic than phenol will release $C{O}_2$ on reaction with $NaHC{O}_3$.

Due to -R effect of $N{O}_2$ it will withdraw the electron density toward themself so the ease of removal of $H^{+}$ is increased so it more acidic than phenol so it will react with sodium bicarbonate to give $C{O}_2$

Benzoic acid is stronger than phenol so it also liberate $C{O}_2$ with $NaHC{O}_3$ solution.

$N{H}_2$ are +R effect group, it increase the electron density on OH and make them less acidic than phenol.

Thus,

enough to liberate $C{O}_2$ with $NaHC{O}_3$ solution.
Hence, option (d) is correct.