$$A$$ and $$B$$ formed in the following reactions are:
$$CrO_2Cl_2 + 4NaOH \rightarrow A + 2NaCl + 2H_2O$$
$$A + 2HCl + 2H_2O_2 \rightarrow B + 3H_2O$$

Chromyl chloride $$CrO_2Cl_2$$ contains chromium in the $$+6$$ oxidation state. In alkaline medium chloride ions are replaced by oxide ions, converting the molecule into the chromate ion $$CrO_4^{2-}$$.

Writing the reaction with hydroxide ions first:
$$CrO_2Cl_2 + 4OH^- \rightarrow CrO_4^{2-} + 2Cl^- + 2H_2O$$

On adding sodium ions to both sides, the molecular equation becomes
$$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$$

Thus, $$A = Na_2CrO_4$$ (sodium chromate).

Next, sodium chromate reacts with dilute acid and hydrogen peroxide. Under mildly acidic conditions the chromate ion forms a blue peroxo complex, commonly written as $$CrO_5$$ (chromium(VI) peroxide):

Ionic form of the reaction:
$$CrO_4^{2-} + 2H^+ + 2H_2O_2 \rightarrow CrO_5 + 3H_2O$$

Adding the spectator sodium and chloride ions gives the required molecular equation:
$$Na_2CrO_4 + 2HCl + 2H_2O_2 \rightarrow CrO_5 + 2NaCl + 3H_2O$$

Hence, $$B = CrO_5$$.

Therefore
$$A = Na_2CrO_4,\qquad B = CrO_5$$, which corresponds to Option A.