Solubility when multiple salts are involved
I'm not sure what your train of thought was on this problem. Maybe annotating the problem would help see where you've gone wrong.

We're given that $K_{\mathrm{sp}}$ = $[\ce{Ag+}][\ce{Cl-}] = 1.6\times10^{-10}~\ce{mol^2/L^2}$. Since the solution already contains a known concentration of the chloride ion, we will solve for the amount of $[\ce{Ag+}]$ that will dissolve from the added $\ce{AgCl}$ to figure out the molar solubility of $\ce{AgCl}$.

$$[\ce{Ag+}] = \frac{K_{\mathrm{sp}}}{[\ce{Cl-}]} = \frac{1.6\times10^{-10}~\mathrm{mol^2/L^2}}{8\times10^{-3}~\mathrm{mol/L}} = 2\times10^{-8}\mathrm{mol/L}$$ 

Note that $[\ce{Cl-}]$ is double that of $\ce{[CaCl2]}$ since $$\ce{CaCl2&lt;=&gt;Ca^2+ + 2Cl-}$$

Since the dissolution of $\ce{AgCl}$ is in a one-to-one ratio with respect to the silver ion, $\ce{Ag+}$, the above answer is sufficient to show the molar solubility of $\ce{AgCl}$ in the given solution.

Any further chloride ions that dissolve will be negligible to the amount that is already in solution (according to the final solubility, only about $2\times10^{-8}~\mathrm{M}$ would be added, which is negligible to $8\times10^{-3}~\mathrm{M}$ by a factor of $400000:1$).