Which of the following compounds is least likely to give effervescence of $$CO_2$$ in presence of aq. NaHCO$$_3$$?

Effervescence of $$CO_2$$ with aqueous $$NaHCO_3$$ is observed only when the organic compound can donate a proton more acidic than the first proton of carbonic acid $$H_2CO_3$$.

The equilibrium that decides the test is
$$\text{HA} + NaHCO_3 \;\rightleftharpoons\; NaA + H_2CO_3 \;\rightleftharpoons\; NaA + CO_2\uparrow + H_2O$$
For the reaction to proceed to the right, the acid strength order must be $$\text{p}K_a(\text{HA}) \lt \text{p}K_a(H_2CO_3)\;( \approx 6.4 ).$$

Now examine each option:

Case A: Benzoic acid, $$C_6H_5COOH$$
$$\text{p}K_a \approx 4.2 \lt 6.4$$  ⇒  reacts with $$NaHCO_3$$, brisk effervescence. Case B: p-Nitrobenzoic acid, $$p\text{-}O_2N\!-\!C_6H_4COOH$$
The strongly -NO2 group withdraws electrons, making the acid still stronger, $$\text{p}K_a \approx 3.4$$. Hence it reacts even more readily with $$NaHCO_3$$. Case C: Salicylic acid (o-hydroxybenzoic acid), $$o\text{-}HO\!-\!C_6H_4COOH$$
Intramolecular H-bonding stabilises its conjugate base and the acid is strong, $$\text{p}K_a \approx 2.97$$. Therefore it also gives effervescence. Case D: p-Cresol (p-methylphenol), $$p\text{-}CH_3\!-\!C_6H_4OH$$
This compound is a phenol, not a carboxylic acid. Phenolic -OH is much less acidic ( $$\text{p}K_a \approx 10$$ ) than carbonic acid, hence $$NaHCO_3$$ cannot deprotonate it. No $$CO_2$$ gets evolved.

Thus the compound least likely to give effervescence of $$CO_2$$ with aqueous $$NaHCO_3$$ is the phenol in Option D.

Final answer: Option D which is: p-Cresol (or any phenolic compound shown in Option D).