The solubility product of $Pb{l}_2$ is $8.0\times {10}^{-9}.$ The solubility of lead iodide in 0.1 molar solution of lead nitrate is 

given ${\left[K_{sp}\right]}_{Pb{l}_2}=8\times {10}^{-8}$

To calculate solubility of $Pb{l}_2$ in 0.1M solution of $Pb{\left(N{O}_3\right)}_2.$

$\left(i\right) Pb{\left(N{O}_3\right)}_2\to P{b}^{2+}\left(aq\right)+2N{O_3}^{-}\left(aq\right)$

(ii)  $\begin{array}{l}0.\overline{1}M0.\overline{1}M0.\overline{2}M\\ Pb{l}_2\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+2l^{-}\left(aq\right)\\ s2s\end{array}$

$\begin{array}{l}\therefore \left[P{b}^{2+}\right]=S+0.1\approx 0.1\\ \because S<<0.1\end{array}$

Now, $K_{sp}=8\times {10}^{-9}$

$\left[P{b}^2\right]{\left[I^{-}\right]}^2=8\times {10}^{-9}$

$0.1\times {\left(2S\right)}^2=8\times {10}^{-9}$

$4S^2=8\times {10}^{-8}\Rightarrow S=\sqrt{2\times {10}^{-8}}M$