When m-chlorobenzaldehyde is treated with 50% KOH solution, the product(s) obtained is (are)

m-Chlorobenzaldehyde, $m-ClC_6H_4CHO$, has no $\alpha$-hydrogen. With concentrated (50%) alkali it undergoes the Cannizzaro reaction (self oxidation-reduction).
One molecule is oxidised to the carboxylate ion and another is reduced to the alcohol:
$2\,m-ClC_6H_4CHO + KOH \rightarrow m-ClC_6H_4COO^-K^+ + m-ClC_6H_4CH_2OH$
The aryl C–Cl bond is not affected by KOH under these conditions (aryl halides are very unreactive towards nucleophilic substitution), so Cl is not replaced by OH.
So the products are the m-chlorobenzoate ion and m-chlorobenzyl alcohol.