Answer
Increases.
Work Step by Step
Because that lead (II) iodide solution is saturated, it can dissolve no more of lead (II) iodide without precipitation. In this case, the reaction quotient, $Q=[Pb^{2+}][I^{-}]^2$, is equal to $Ksp$.
When solid $KI$ is added to this solution, it dissociates to form $K^{+}$ and $I^{-}$. The concentration of $I^{-}$ rises, making $[I^{-}]^2$ rises, thus rising the reaction quotient $Q$. Now, $Q$ is certainly larger than $Ksp$.
When $Q>Ksp$, the reaction proceeds to the other way, making precipitate so that the ions' concentrations would be lowered and equal $Ksp$ again. Therefore, when we add $KI$, the amount of $PbI_{2}$ solid increases.
