Identify the metal halide. The starting halide must give a white precipitate with NaOH and contain a halide ion that can be oxidised to a halogen which turns starch-iodide paper blue. This points to ${\text{MnCl}}_2$:

$${\text{MnCl}}_2+2\text{NaOH}⟶\underset{P,\text{white}}{\underbrace{{\text{Mn(OH)}}_2}}+2\text{NaCl}.$$

So the filtrate Q contains chloride ions.

Formation of X. Manganese(II) hydroxide is oxidised by PbO${}_2$ in acidic medium (analogous to the Mn(II) to permanganate test) to the deep purple permanganate ion: $$2{\text{Mn(OH)}}_2+5{\text{PbO}}_2+6{\text{H}}^{+}⟶2{\text{MnO}}_4^{-}+5{\text{Pb}}^{2+}+6{\text{H}}_2\text{O}(\text{simplified}).$$ Thus X = MnO${}_4^{-}$.

Formation of Y. Chloride in conc. H${}_2$SO${}_4$ with MnO(OH)${}_2$ as oxidant gives chlorine gas: $${\text{Cl}}^{-}\stackrel{{\text{MnO(OH)}}_2,{\text{conc. H}}_2{\text{SO}}_4}{\to }{\text{Cl}}_2.$$ Chlorine oxidises iodide on the starch-iodide paper to iodine, giving the characteristic blue colour with starch: ${\text{Cl}}_2+2{\text{I}}^{-}\to 2{\text{Cl}}^{-}+{\text{I}}_2$.

Hence Y = Cl${}_2$. Answer: (C).