Which one of the following has a square planar geometry?

${{}_{27}\mathrm{Co}}^{2+}-1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{7}4{\mathrm{s}}^0$

As ${\mathrm{Cl}}^{-}$is weak field ligand so no pairing up. Hence it is ${\mathrm{sp}}^3$ hybridized giving tetrahedral geometry.

${\mathrm{Fe}}^{2+}-1{ \mathrm{s}}^{2}2{ \mathrm{s}}^{2}2{\mathrm{p}}^{6}3{ \mathrm{s}}^{2}3{\mathrm{p}}^{6}3{ \mathrm{d}}^{6}4{ \mathrm{s}}^8$

Due to ${\mathrm{Cl}}^{-}$, back pairing is not observed so it will be ${\mathrm{sp}}^3$ hybridized giving tetrahedral geometry.

${\mathrm{Ni}}^{2+}-1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{8}4{\mathrm{s}}^0$

Because weak ligand, back pairing is not observed so it will be ${\mathrm{sp}}^3$ i.e. tetrahedral geometry. 

All the complexes of ${\mathrm{Pt}}^{2+}$ are square planar including those with weak field ligand such as halide ions thus (4) is correct.