The OP asked the question again, and proposed the following solution from his book.
My textbook says to solve it in the following way:
1) Ignore the 0.10 M KI since it doesn't matter.
2) Since you have excess AgCl(s), you can calculate the Ag+ concentration using its 𝐾sp.
𝐾sp=1.77×10−10=𝑥^2
𝑥=1.3×10^−5
This is the Ag+ concentration
3) Substitute calculated Ag+ concentration into equilibrium for 𝐾sp of AgI.
𝐾sp=8.51×10^−17=[Ag+][I−]
8.51×10^−17=[1.3×10^−5][I−]
[I−]=6.4×10^−12𝑀
The book answer is wrong!!
The crux of the matter is what limits the final concentration of silver. Point (2) in the book answer is wrong, which makes point (1) wrong also.
The equilibrium for the $\ce{AgCl}$ precipitate depends on both the concentrations of both $\ce{Ag+}$ and $\ce{Cl-}$. (We're ignoring the activity/concentration factor for the ions.)
Given the reaction:
$$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)}\tag{1}$$
The reaction is an exchange reaction. The AgCl(s) dissolves via:
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}\tag{2}$$
and then KI precipitates via:
$$\ce{Ag+(aq) + I-(aq) <=> KI(s)}\tag{3}$$
When the reaction ends, then the $\ce{K+}$ cations left in solution must be balanced by the same concentration of anions in the solution. The anions in this case could be either $\ce{Cl-}$ or $\ce{I-}$. However due to the much greater insolubility of $\ce{AgI}$ we know that $\ce{[Cl-] \gg [I-]}$. So the solution must end up (to two significant figures), $\ce{\pu{0.10 M} K+}$ and $\ce{\pu{0.10 M} Cl-}$ with traces of $\ce{Ag+}$ and $\ce{I-}$.
For the book answer the final concentration of $\ce{Ag+}$ is $\ce{1.3\times 10^{−5}}$. Since the final $\ce{Cl-}$ is $\pu{0.10 M}$ that means that
$$\mathrm{K'_{sp,AgCl}} =\ce{[Ag+][Cl-]}= (1.3\times 10^{−5})(0.1) = 1.3\times 10^{−6}$$
The acknowledged value for $\mathrm{K_{sp,AgCl}} = \pu{1.77\times 10^{-10}}$. Thus Point (2) in the book answer must be wrong.