The solubility product of AgI at $25^\circ C$ is $1.0 \times 10^{-16}$ mol$^2$ L$^{-2}$. The solubility of AgI in $10^{-4}$ N solution of KI at $25^\circ C$ is approximately (in mol L$^{-1}$) :

KI is fully ionised and supplies $[I^-] = 10^{-4}$ M (for KI, normality = molarity).
Let the solubility of AgI in this solution be s. Then $[Ag^+] = s$ and $[I^-] = s + 10^{-4} \approx 10^{-4}$ (since s is very small).
$K_{sp} = [Ag^+][I^-] = s \times 10^{-4}$
$s = \dfrac{1.0 \times 10^{-16}}{10^{-4}} = 1.0 \times 10^{-12}$ mol L$^{-1}$
(In pure water the solubility would be $\sqrt{10^{-16}} = 10^{-8}$ mol L$^{-1}$; the common ion $I^-$ reduces it.)