Explanation

(d)
$Pb(OH)_{2} \rightleftharpoons Pb^{2 +} + 2OH^{-}\$
$K_{sp} = \left\lbrack Pb^{2 +} \right\rbrack\left\lbrack OH^{-} \right\rbrack^{2} = S \times (2S)^{2}$
$K_{sp} = 4s^{3} = 4 \times \left( 6.7 \times 10^{- 6} \right)^{3}$
$\ = 1.20 \times 10^{- 15}$
In a solution with pH=8
$\left\lbrack H^{+} \right\rbrack = 10^{- 8}\ \$
$and\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrack {OH}^{-} \right\rbrack = 10^{- 6}$
$1.20 \times 10^{- 15} = \left\lbrack Pb^{2 +} \right\rbrack\left\lbrack 10^{- 6} \right\rbrack^{2}$
$\left\lbrack Pb^{2 +} \right\rbrack = \frac{1.2 \times 10^{- 15}}{\left\lbrack 10^{- 6} \right\rbrack^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.2 \times 10^{- 3}$M