A hydrated solid X on heating initially gives a monohydrated compound Y. Y upon heating above 373 K leads to an anhydrous white powder Z. X and Z, respectively, are;

We start by translating the verbal information into chemical language. The solid $$X$$ is said to be a hydrated compound, i.e. it contains water of crystallisation. When $$X$$ is heated, it first loses water until only a monohydrate $$Y$$ remains. Mathematically:

$$X \;{\xrightarrow{\text{mild heat}}}\; Y + \text{water vapour}$$

Again heating $$Y$$ above $$373\ \text{K}$$ removes its last molecule of water and produces an anhydrous white powder $$Z$$:

$$Y \;{\xrightarrow{T > 373\ \text{K} }}\; Z +$$ water vapour

The two-step loss of water strongly hints at sodium carbonate, because:

Formula for washing soda (decahydrate): $$\mathrm{Na_2CO_3\cdot 10H_2O}$$.

When gently heated (just above room temperature but below $$373\ \text{K}$$), washing soda loses $$9$$ of its $$10$$ water molecules and becomes the monohydrate:

$$\mathrm{Na_2CO_3\cdot 10H_2O}\;\;{\xrightarrow{\text{heat}}}\;\;\mathrm{Na_2CO_3\cdot H_2O} + 9\mathrm{H_2O}$$

Further heating the monohydrate above $$373\ \text{K}$$ drives off the last water molecule, giving anhydrous sodium carbonate, commonly called soda ash:

$$\mathrm{Na_2CO_3\cdot H_2O}\;\;{\xrightarrow{T > 373\ \text{K}}}\;\;\mathrm{Na_2CO_3} + \mathrm{H_2O}$$

Thus we can match each species with the symbols used in the question:

$$X = \mathrm{Na_2CO_3\cdot 10H_2O}\quad(\text{washing soda})$$

$$Y = \mathrm{Na_2CO_3\cdot H_2O}\quad(\text{monohydrate})$$

$$Z = \mathrm{Na_2CO_3}\quad($$ soda ash, a white anhydrous powder $$)$$

Comparing with the options, only choice A lists washing soda for $$X$$ and soda ash for $$Z$$.

Hence, the correct answer is Option A.