Compound A gives blue colour solution with excess of ${\mathrm{NH}}_4\mathrm{OH}$, it gives chocolate brown colour with ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$. Complex A is:

${\mathrm{CuSO}}_4$ reacts with  ${\mathrm{NH}}_4\mathrm{OH}$ to form a deep blue copper ammonium sulphate complex. 

${\mathrm{CuSO}}_4+4{\mathrm{NH}}_4\mathrm{OH}⟶\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{SO}}_4+4{\mathrm{H}}_2\mathrm{O}$

When ${\mathrm{CuSO}}_4$ solution is added to ${\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$ the product formed has the formula ${\mathrm{Cu}}_2\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$.

$2{\mathrm{CuSO}}_4+{\mathrm{K}}_4\left(\mathrm{Fe}(\mathrm{CN}{)}_6\right]\to {\mathrm{Cu}}_2\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]+2{\mathrm{K}}_2{\mathrm{SO}}_4$

Hence option A is correct.