The compound $\mathrm{PtCl}{}_4\cdot 4\mathrm{NH}{}_3$ (Pt coordination number 6) gives 2 mol of $\mathrm{AgCl}$ per mole with excess $\mathrm{AgNO}{}_3$. Its correct formulation is

Two precipitable chlorides means two free $\mathrm{Cl}{}^{-}$ ions, so two chlorides are inside. Coordination number $=4+2=6$, matching Pt(IV): the cation is $+2$, balanced by two $\mathrm{Cl}{}^{-}$. The option with 3 chlorides inside would have coordination number 7 and give only 1 AgCl.