In the above reaction, product B is

The reactant is phenol, $$C_6H_5OH$$. In the first step it is treated with aqueous $$NaOH$$ followed by dry $$CO_2$$ at about $$400\text{ K}$$ and $$4 \text{-} 7 \text{ atm}$$ pressure. This is the Kolbe-Schmitt reaction.

Kolbe-Schmitt reaction principle
Under the above conditions the phenoxide ion $$C_6H_5O^-$$ attacks $$CO_2$$, and an $$-COO^-$$ group is introduced preferentially at the ortho position (because of intramolecular H-bond stabilisation in the product). The intermediate formed is sodium salicylate, $$o\text{-}HO\,C_6H_4COO^-Na^+$$.

On subsequent acidification with dilute $$HCl$$ (or any mineral acid) the sodium salt is protonated to give the free acid.

Therefore, product B obtained after the acidification step is ortho-hydroxybenzoic acid (commonly called salicylic acid), having the structure
$$HO-\underset{1}{C_6H_4}-COOH$$ with the $$-OH$$ and $$-COOH$$ groups in the ortho (1,2) positions.

Among the given choices, Option A corresponds to this structure. Hence, the correct answer is Option A.