This problem involves a systematic qualitative analysis of a mixture of two inorganic salts. We need to identify the cation present in precipitate (P) based on a series of chemical tests.

Step 1: Analysis of the reaction with dilute HCl

The first step is the addition of dilute $HCl$ to the aqueous solution of the salt mixture. This treatment results in a precipitate (P) and a filtrate (Q).

$\text{Salt Mixture (aq)} + \text{dilute } HCl \rightarrow \text{Precipitate (P)} \downarrow + \text{Filtrate (Q)}$

Addition of dilute $HCl$ is the group test for Group 1 cations in qualitative analysis. The cations that precipitate as chlorides are $Ag^+$, $Pb^{2+}$, and $Hg_2^{2+}$.

• $Ag^+ (aq) + Cl^- (aq) \rightarrow AgCl(s)$ (white precipitate)
• $Pb^{2+} (aq) + 2Cl^- (aq) \rightarrow PbCl_2(s)$ (white precipitate)
• $Hg_2^{2+} (aq) + 2Cl^- (aq) \rightarrow Hg_2Cl_2(s)$ (white precipitate)

Therefore, the precipitate (P) must contain one of these three cations. This eliminates option D ($Hg^{2+}$), which belongs to Group 2 and does not precipitate with dilute $HCl$.

Step 2: Analysis of the solubility of Precipitate (P)

The problem states that the precipitate (P) dissolves in hot water. We can use this information to distinguish between the possible Group 1 chlorides:

• $AgCl$ (Silver chloride) is insoluble in hot water. It is soluble in aqueous ammonia.
• $PbCl_2$ (Lead(II) chloride) is sparingly soluble in cold water but is soluble in hot water.
• $Hg_2Cl_2$ (Mercury(I) chloride) is insoluble in hot water. It reacts with aqueous ammonia to form a black precipitate.

Since precipitate (P) dissolves in hot water, it must be $PbCl_2$. Therefore, the cation present in precipitate (P) is $Pb^{2+}$.

Step 3: Confirmation from the analysis of Filtrate (Q) (for completeness)

Although we have already identified the cation in P, let's analyze the information about filtrate Q to ensure consistency.

1. Filtrate (Q) + $H_2S$ in dilute acid medium $\rightarrow$ No precipitate. This is the test for Group 2 cations. The absence of a precipitate indicates that cations like $Cu^{2+}$, $Cd^{2+}$, $As^{3+}$, etc., are absent. (Note: $Pb^{2+}$ is also in Group 2, but most of it was precipitated as $PbCl_2$. Any residual $Pb^{2+}$ would precipitate as $PbS$ if its concentration were high enough).

2. Filtrate (Q) + $H_2S$ in ammoniacal medium $\rightarrow$ Precipitate (R). This is the test for Group 3 and 4 cations. An ammoniacal medium ($NH_4OH/NH_4Cl$ buffer) would first precipitate Group 3 cations as hydroxides (e.g., $Al(OH)_3$, $Fe(OH)_3$, $Cr(OH)_3$). Then, passing $H_2S$ would precipitate Group 4 cations as sulfides (e.g., $ZnS$, $MnS$, $NiS$, $CoS$).

3. Precipitate (R) + $H_2O_2$ in aqueous $NaOH$ $\rightarrow$ Coloured solution (S). This is a characteristic test for $Cr^{3+}$. The precipitate (R) would be chromium(III) hydroxide, $Cr(OH)_3$ (greenish). This precipitate is oxidized by hydrogen peroxide in an alkaline medium to form the soluble, yellow chromate ion ($CrO_4^{2-}$). $2Cr(OH)_3 + 3H_2O_2 + 4OH^- \rightarrow 2CrO_4^{2-} + 8H_2O$ The resulting yellow solution (S) is due to the chromate ion. This indicates the second cation in the mixture was $Cr^{3+}$.

Conclusion

Based on the fact that the precipitate (P) formed with dilute $HCl$ is soluble in hot water, the cation in P must be $Pb^{2+}$.

• Option A: $Pb^{2+}$ forms $PbCl_2$, which is soluble in hot water. This is correct.
• Option B: $Hg_2^{2+}$ forms $Hg_2Cl_2$, which is insoluble in hot water.
• Option C: $Ag^+$ forms $AgCl$, which is insoluble in hot water.
• Option D: $Hg^{2+}$ does not precipitate with dilute $HCl$.

Thus, the precipitate P contains $Pb^{2+}$.