Thermal decomposition of a Mn compound (X) at 513 K results in compound Y, MnO$$_2$$ and a gaseous product. MnO$$_2$$ reacts with NaCl and concentrated H$$_2$$SO$$_4$$ to give a pungent gas Z. X, Y and Z, respectively, are:

We have to identify the manganese compound $$X$$ which, on heating at $$513\ \text{K}$$, gives another manganese compound $$Y$$, $$\text{MnO}_2$$ and a gaseous product. The classical thermal decomposition equation for the dark-purple permanganate salt is stated first:

$$\text{Thermal decomposition formula:}\qquad 2\ \text{KMnO}_4 \;\xrightarrow{\,\Delta ,\ 513\ \text{K}\,}\; \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2$$

Now, applying this equation step by step:

On the left side we begin with $$2\ \text{KMnO}_4$$. On the right side the potassium atoms must still be two in number, so we obtain $$\text{K}_2\text{MnO}_4$$. One of the two manganese atoms is already used in $$\text{K}_2\text{MnO}_4$$, leaving the second manganese atom to appear as $$\text{MnO}_2$$. Oxygen balance then requires one molecule of $$\text{O}_2$$ gas. Thus every atom remains conserved, and the gaseous product is clearly $$\text{O}_2$$.

Hence, by direct comparison with the question, we identify

$$X = \text{KMnO}_4$$, $$Y = \text{K}_2\text{MnO}_4$$, gas from decomposition = $$\text{O}_2$$.

Next we use the information that the $$\text{MnO}_2$$ formed further reacts with $$\text{NaCl}$$ and concentrated $$\text{H}_2\text{SO}_4$$ to liberate a pungent gas $$Z$$. We first state the well-known laboratory chlorine preparation equation:

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

In the actual setup, concentrated $$\text{H}_2\text{SO}_4$$ protonates $$\text{NaCl}$$, generating $$\text{HCl}$$ in situ, so the net result is still the evolution of $$\text{Cl}_2$$. Chlorine has the characteristic choking, pungent smell mentioned in the problem. Therefore

$$Z = \text{Cl}_2.$$

Putting all the identifications together we obtain

$$X = \text{KMnO}_4,\quad Y = \text{K}_2\text{MnO}_4,\quad Z = \text{Cl}_2.$$

These correspond exactly to Option D in the given list.

Hence, the correct answer is Option D.