1. We need the increasing order of solubility product, $K_{sp}$ for: $\mathrm{Ca(OH)_2},\ \mathrm{AgBr},\ \mathrm{PbS},\ \mathrm{HgS}$

2. Recall standard relative magnitudes of $K_{sp}$

   • Sulfides of heavy metals are extremely insoluble.
   • Among the given sulfides: $K_{sp}(\mathrm{HgS}) < K_{sp}(\mathrm{PbS})$ because $\mathrm{HgS}$ is one of the least soluble ionic sulfides.
   • Silver bromide is sparingly soluble, but much more soluble than these sulfides: $K_{sp}(\mathrm{AgBr}) > K_{sp}(\mathrm{PbS}),\ K_{sp}(\mathrm{HgS})$
   • Calcium hydroxide is comparatively much more soluble, so it has the largest $K_{sp}$ among these.

3. Thus the increasing order is $\mathrm{HgS} < \mathrm{PbS} < \mathrm{AgBr} < \mathrm{Ca(OH)_2}$

4. Match with options This corresponds to Option C.

5. Answer check with stored correct answer

   • Derived answer: C
   • Stored correct answer: C
   • They agree.

Final Answer: $\boxed{\mathrm{HgS} < \mathrm{PbS} < \mathrm{AgBr} < \mathrm{Ca(OH)_2}}$