When mercuric iodide is added to the aqueous solution of potassium iodide

$\mathrm{HgI}$ reacts with $\mathrm{KI}$, to form ${\mathrm{K}}_2\left[{\mathrm{HgI}}_4\right]$. This causes the amount of ions to decline, which lowers the Van't Hoff factor.
$\Delta {\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{k}}_{\mathrm{f}}\times \mathrm{m}$

Because 'i' has gotten smaller, the freezing point is less depressed, which causes it to rise.

Therefore, option A) Freezing point is raised is the necessary answer.