The number of geometrical isomers of the square planar complex $[\mathrm{Pt}(\mathrm{NH}{}_3)(\mathrm{py})\mathrm{ClBr}]$ (py = pyridine, monodentate) is

For square planar $\mathrm{Mabcd}$, fix a at one corner. Then any one of b, c, d may be trans to a, and the remaining two go to the other two corners as one arrangement. That gives 3 geometrical isomers.