When a salt reacts with $$\mathrm{MnO}_2$$ and concentrated $$\mathrm{H}_2 \mathrm{SO}_4$$, the type of gas evolved depends on the anion present in the salt. In this case, a greenish yellow gas is evolved, which indicates the formation of chlorine gas ($$\mathrm{Cl}_2$$). $$\mathrm{Cl}_2$$ gas is greenish-yellow in color and is typically produced from halide salts (namely chlorides, bromides, or iodides) when they are oxidized. Among the given options, we need to identify a salt that contains a halide which can produce $$\mathrm{Cl}_2$$ upon reaction with $$\mathrm{MnO}_2$$ and concentrated $$\mathrm{H}_2 \mathrm{SO}_4$$.

Option A ($$\mathrm{NH}_4 \mathrm{Cl}$$) contains chloride ions, and when heated with $$\mathrm{MnO}_2$$ and concentrated $$\mathrm{H}_2 \mathrm{SO}_4$$, it can undergo a redox reaction where the chloride ions are oxidized to $$\mathrm{Cl}_2$$ gas. The corresponding reaction can be represented as follows:

$$\mathrm{MnO}_2 + 4\mathrm{HCl} \longrightarrow \mathrm{MnCl}_2 + \mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O}$$

Option B ($$\mathrm{CaI}_2$$) contains iodide ions, which would lead to the liberation of iodine or $$\mathrm{I}_2$$, not a greenish yellow gas.

Option C ($$\mathrm{KNO}_3$$) contains nitrate ions and does not produce a halogen gas upon reaction with $$\mathrm{MnO}_2$$ and $$\mathrm{H}_2 \mathrm{SO}_4$$.

Option D ($$\mathrm{NaBr}$$) contains bromide ions which would lead to the formation of bromine ($$\mathrm{Br}_2$$), a reddish-brown gas, not greenish yellow.

Therefore, the correct answer is Option A ($$\mathrm{NH}_4 \mathrm{Cl}$$), since it is the salt that can produce a greenish yellow gas ($$\mathrm{Cl}_2$$) when reacted with $$\mathrm{MnO}_2$$ and concentrated $$\mathrm{H}_2 \mathrm{SO}_4$$.