Explanation

(c)
$K_{sp}\ of\ AgCl = (solubility\ of\ AgCl)^{2}$
$= \left( 1 \times 10^{- 5} \right)^{2} = 1 \times 10^{- 10}$
Suppose its solubility in 0.1 M $NaCl\ is\ x\ \frac{mol}{L}$
$AgCl \rightleftharpoons {Ag}^{+} + {Cl}^{-}$
$x$ $x$
$NaCl \rightleftharpoons {Na}^{+} + {Cl}^{-}$
0.1 M 0.1 M
$\left\lbrack Cl^{-} \right\rbrack = (x + 0.1)$ M
$K_{sp}\ of\ AgCl = {\lbrack Ag}^{+}\rbrack\lbrack{Cl}^{-}\rbrack$
$= x \times (x + 0.1)$
$1 \times 10^{- 10} = x^{2} + 0.1x$
Higher power of $x$ are neglecated
$1 \times 10^{- 10} = 0.1x$
$x = 1 \times 10^{- 9}$ M