In the following reaction, the product A is

The correct answer is Option (3) → o-Bromobenzoic acid

In reaction 1 Br adds at the ortho position as para position is blocked by $-{NO_2}$ resulting in o-bromotoluene.

In reaction 2 Sn/HCl reduces $-{NO_2}$ group to $-{NH_2}$ group (reduction) 

In third reaction then NaNO₂/HCl forms diazonium salt (diazotization). 

In 4rth reaction $H_3PO_2$ replaces $-{N_2}^+$ with H (removes the diazonium group)

In last reaction $KMnO_4$ oxidizes the methyl to –COOH, yielding ortho-bromobenzoic acid