In the following reaction, $\mathrm{AgCl}+\mathrm{KI}\rightleftharpoons \mathrm{KCl}+\mathrm{AgI}$ As KI is added, the equilibrium is shifted towards right giving more AgI precipitate, because

Both AgCl and AgI are sparingly soluble

The ${\mathrm{K}}_{\mathrm{Sp}} \mathrm{of} \mathrm{AgI}$ is lower than ${\mathrm{K}}_{\mathrm{Sp}} \mathrm{of} \mathrm{AgCl}$

$\mathrm{The} {\mathrm{K}}_{\mathrm{Sp}}\mathrm{of} \mathrm{Ag} \mathrm{I} \mathrm{is} \mathrm{higher} \mathrm{than} {\mathrm{K}}_{\mathrm{Sp}} \mathrm{of} \mathrm{AgCl}$

Both AgCl and AgI have same solubility product