Assume $[\mathrm{Pt}(\mathrm{NH}{}_3){}_2(\mathrm{NO}{}_2){}_2]$ is square planar and each nitrite ligand can bind through N or through O. The total number of distinct isomers (geometrical and linkage) of the neutral complex is

Both nitrite ligands N-bound gives cis and trans (2). Both O-bound gives cis and trans (2). One N-bound and one O-bound again gives cis and trans arrangements of the two $\mathrm{NH}{}_3$ groups (2), because the two different nitrite forms can still lie adjacent or opposite. Total $=2+2+2=6$.