The standard reduction potential for ${\mathrm{Cu}}^{2+}/\mathrm{Cu}$ is +0.34V. If  ${\mathrm{K}}_{\mathrm{sp}}$ of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$ is $1.0 \times {10}^{-19}$, the reduction potential at pH =14 for the above couple is

Here we have ${\mathrm{K}}_{\mathrm{sp}}$ of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2=1.0\times {10}^{-19}$
Here ${\mathrm{K}}_{\mathrm{sp}}=\mathrm{solubility} \mathrm{product}$
Standard reduction potential for ${\mathrm{Cu}}^{2+}/\mathrm{Cu}=+0.34\mathrm{V}$
What will be the reduction potential at pH=14
pH is given by 
$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=14 \mathrm{or} \left[{\mathrm{H}}^{+}\right]={10}^{-4}$
Hence $\left[{\mathrm{OH}}^{-}\right]=1\mathrm{M}\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={10}^{-4}$
For $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$
$\mathrm{Cu}{\left(\mathrm{OH}\right)}_2\to {\mathrm{Cu}}^{2+}+2{\mathrm{OH}}^{-}$
Now ${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Cu}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2$
For the reaction ${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}, \mathrm{n}=2$
Therefore, 

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.0591}{\mathrm{n}}\log \frac{1}{\left[{\mathrm{Cu}}^{2+}\right]}$

By substituting above values

${\mathrm{E}}_{\mathrm{cell}}=0.34-\frac{0.0591}{2}\log \frac{1}{1\times {10}^{=19}}$

${\mathrm{E}}_{\mathrm{cell}}=-0.22\mathrm{V}$

Hence option D is correct.