All of the given cations are group -1 cations, and when HCl is added to their aqueous solution, they will precipitate their corresponding chlorides

$\begin{array}{l}{\mathrm{Ag}}^{+}\left(aq\right)+\mathrm{HCl}\left(aq\right)\to \mathrm{AgCl}\left(s\right)+{\mathrm{H}}^{+}\left(aq\right)\\ {\mathrm{Cu}}^{2+}+\left(aq\right)+2\mathrm{HCl}\left(aq\right)\to {\mathrm{CuCl}}_2\left(s\right)+2{\mathrm{H}}^{+}\left(aq\right)\\ {\mathrm{Hg}}_2^{2+}+\left(aq\right)+2\mathrm{HCl}\left(aq\right)\to {\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(s\right)+2{\mathrm{H}}^{+}\left(aq\right)\\ {\mathrm{Pb}}^{2+}+\left(aq\right)+2\mathrm{HCl}\left(aq\right)\to {\mathrm{PbCl}}_2\left(s\right)+2{\mathrm{H}}^{+}\left(aq\right)\end{array}$

All the precipitates are insoluble in cold water but out of all these only lead (II) chloride is soluble in warm water.

Hence the correct option is D