${\mathrm{CuSO}}_4$ solution reacts with excess KCN to give:

${\mathrm{Cu}}^{2+}$ is first converted to ${\mathrm{Cu}}^{+}$ and then precipitated as a result of adding KCN to a solution of ${\mathrm{CuSO}}_4$.
This interacts with extra KCN to create a soluble ${\mathrm{K}}_3\left[\mathrm{Cu}(\mathrm{CN}{)}_4\right]$ complex.

The following reaction takes place when calcium sulphate solution reacts with excess KCN 

$\begin{array}{l}{\mathrm{CuSO}}_4+2\mathrm{KCN}\to \mathrm{Cu}(\mathrm{CN}{)}_2+{\mathrm{K}}_2{\mathrm{SO}}_4\\ 2\mathrm{Cu}(\mathrm{CN}{)}_2\to 2\mathrm{CuCN}+(\mathrm{CN}{)}_2\\ \mathrm{CuCN}+3\mathrm{KCN}\to {\mathrm{K}}_3\mathrm{Cu}(\mathrm{CN}{)}_4\end{array}$

Hence, option D is the correct answer.