$NH _{3}$ is neutral, making the first complex positively charged overall.
$Cl$ has a $-1$ charge, making the second complex the anion and so platinum ends with -ate.
Therefore, the complex with $NH _{3}$ comes first, followed by the one with $Cl$.
Option (A) is correct. The I.U.P.A.C name for $\left[ Pt \left( NH _{3}\right)_{4}\right]\left[ PtCl _{4}\right]$
is tetraammineplatinum(II) tetrachloridoplatinate(II)