This is a qualitative analysis problem to identify two metal cations, X and Y, from a series of chemical tests.

1. Initial Information: We start with a colourless aqueous solution containing the nitrates of two metals, X and Y. Nitrates are generally soluble in water. The colourless nature of the solution suggests that the metal ions are not from the transition elements that typically form coloured ions (like $Cu^{2+}$, $Fe^{2+}$, $Fe^{3+}$, $Co^{2+}$, $Ni^{2+}$, etc.).

2. Reaction with NaCl: When an aqueous solution of $NaCl$ is added, a white precipitate is formed. This indicates that at least one of the metal cations forms an insoluble chloride. The common metal cations that form white precipitates with chloride ions ($Cl^−$) are $Ag^+$, $Pb^{2+}$, and $Hg_2^{2+}$.

   • $Ag^+ + Cl^- \rightarrow AgCl(s)$ (white precipitate)
   • $Pb^{2+} + 2Cl^- \rightarrow PbCl_2(s)$ (white precipitate)
   • Let's examine the options: Ag, Pb, Cd, Zn.
   • $CdCl_2$ and $ZnCl_2$ are soluble in water. Therefore, metals Cd and Zn can be ruled out as the ones forming the precipitate. This eliminates options B, C, and D. The metals must be Ag and Pb.

3. Solubility in Hot Water: The white precipitate is found to be partly soluble in hot water. This gives us more information to distinguish between the possible chloride precipitates.

   • $AgCl$ is insoluble in both cold and hot water.
   • $PbCl_2$ is sparingly soluble in cold water but its solubility increases significantly in hot water.
   • Since the precipitate is partly soluble, it must be a mixture of a soluble and an insoluble component in hot water. This confirms the precipitate is a mixture of $AgCl$ and $PbCl_2$.
   • The part that dissolves in hot water is $PbCl_2$, forming solution Q.
   • The part that remains as a solid residue is $AgCl$, which is residue P.

4. Analysis of Residue P: The residue P is $AgCl$. The problem states that P is soluble in aqueous $NH_3$ and also in excess sodium thiosulphate.

   • Reaction with aqueous $NH_3$: $AgCl$ reacts with ammonia to form a soluble complex, diamminesilver(I) chloride. $AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+Cl^-(aq)$
   • Reaction with sodium thiosulphate ($Na_2S_2O_3$): $AgCl$ reacts with thiosulphate ions to form another soluble complex, dithiosulphatoargentate(I) ion. $AgCl(s) + 2S_2O_3^{2-}(aq) \rightarrow [Ag(S_2O_3)_2]^{3-}(aq) + Cl^-(aq)$
   • These reactions are characteristic of $AgCl$ and confirm that one of the metals is Silver (Ag).

5. Analysis of Solution Q: Solution Q contains the chloride that dissolved in hot water, which is $PbCl_2$. Thus, solution Q contains $Pb^{2+}$ ions. The problem states that solution Q gives a yellow precipitate with $KI$.

   • Reaction with $KI$: $Pb^{2+}$ ions react with iodide ions ($I^−$) to form lead(II) iodide, which is a characteristic bright yellow precipitate. $Pb^{2+}(aq) + 2I^-(aq) \rightarrow PbI_2(s)$
   • This test confirms that the other metal is Lead (Pb).

6. Conclusion: Based on the step-by-step analysis of the reactions, the two metals, X and Y, are Silver (Ag) and Lead (Pb). Therefore, the correct option is A.