Write the dissolution equilibrium for silver chloride (AgCl).

The ion product expression for AgCl is $Q_{sp} = [\text{Ag}^+][\text{Cl}^-]$.
Given $[\text{Ag}^+] = 1.0 \times 10^{-3} \text{ M}$ and $[\text{Cl}^-] = 1.0 \times 10^{-3} \text{ M}$.

Given $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$.
We have $Q_{sp} = 1.0 \times 10^{-6}$ and $K_{sp} = 1.8 \times 10^{-10}$.
Since $1.0 \times 10^{-6} > 1.8 \times 10^{-10}$, $Q_{sp} > K_{sp}$.

When $Q_{sp} > K_{sp}$, the solution is supersaturated, and a precipitate will form to reach equilibrium.
Therefore, a precipitate of AgCl will form.

The final answer is $\boxed{\text{A precipitate will form}}$.

:

Given $K_{sp}(\text{PbI}_2) = 8.5 \times 10^{-9}$.

The solution contains $0.10 \text{ M KI}$. Since KI is a strong electrolyte, it dissociates completely: $\text{KI}(aq) \to \text{K}^+(aq) + \text{I}^-(aq)$.
So, the initial concentration of $\text{I}^-$ from KI is $0.10 \text{ M}$.
Let 's' be the molar solubility of PbI$_2$.

Since $K_{sp}$ is very small ($8.5 \times 10^{-9}$), we can assume that $2s$ is much smaller than $0.10 \text{ M}$.
Therefore, $0.10 + 2s \approx 0.10$.

The assumption that $2s \ll 0.10$ is valid because $2 \times (8.5 \times 10^{-7}) = 1.7 \times 10^{-6}$, which is indeed much smaller than 0.10.

The final answer is $\boxed{8.5 \times 10^{-7} \text{ M}}$.