The one that is not expected to show isomerism is:

First we write the four species once more so that we can examine them calmly:

$$[Ni(NH_3)_4(H_2O)_2]^{2+},\;\;[Ni(en)_3]^{2+},\;\;[Ni(NH_3)_2Cl_2],\;\;[Pt(NH_3)_2Cl_2]$$

All the metal ions are in the $$+2$$ oxidation state, so the electronic configurations are $$Ni^{2+}\;(3d^8)$$ and $$Pt^{2+}\;(5d^8)$$. Complexes of $$d^8$$ ions are usually square-planar when the ligands are of moderate or strong field strength and tetrahedral when the field is weak. With this background we now analyse each option one by one.

We have a useful fact: an octahedral complex of the form $$MA_4B_2$$ always gives two geometrical isomers (cis and trans), while a square-planar complex of the form $$MA_2B_2$$ also gives two geometrical isomers. On the other hand a tetrahedral complex of the form $$MA_2B_2$$ gives only one arrangement, because in a tetrahedron every position is equivalent to every other position by rotation.

Now let us consider each complex.

For $$[Ni(NH_3)_4(H_2O)_2]^{2+}$$ the coordination number is $$6$$, so the geometry is octahedral. The formula type is $$MA_4B_2$$ (four $$NH_3$$ and two $$H_2O$$). Using the rule quoted above, the complex can exist as a cis and a trans form, so geometrical isomerism is expected.

For $$[Ni(en)_3]^{2+}$$ the metal is again octahedral, but all three ligands are identical bidentate ligands $$(en=\;H_2N-CH_2-CH_2-NH_2)$$. An octahedral complex containing three bidentate ligands of the same type is well known to be chiral; the two mirror-image forms are called $$\Delta$$ and $$\Lambda$$ isomers. Hence this complex shows optical isomerism.

For $$[Pt(NH_3)_2Cl_2]$$ the metal $$Pt^{2+}$$ is almost always square-planar because the $$5d$$ orbitals experience a strong ligand field. The coordination number is $$4$$ and the formula type is $$MA_2B_2$$, so two geometrical isomers (cis and trans) are possible. Indeed, the cis isomer is the famous anti-cancer drug “cis-platin”.

Finally we reach $$[Ni(NH_3)_2Cl_2]$$. Here the ligands $$NH_3$$ (moderate field) and $$Cl^-$$ (weak field) do not force $$Ni^{2+}$$ into the square-planar arrangement; instead the ion usually adopts the tetrahedral geometry because that costs less pairing energy for a $$3d^8$$ metal with such ligands. The coordination number is $$4$$ and the tetrahedral formula type is again $$MA_2B_2$$. As stated earlier, in a tetrahedral complex every vertex is equivalent by simple rotation, so only one spatial arrangement is possible. There is therefore no geometrical or optical isomerism expected for this complex.

Putting all the observations together:

$$[Ni(NH_3)_4(H_2O)_2]^{2+}$$ → shows cis/trans geometrical isomerism.

$$[Ni(en)_3]^{2+}$$ → shows optical isomerism ($$\Delta,\;\Lambda$$).

$$[Pt(NH_3)_2Cl_2]$$ → shows cis/trans geometrical isomerism.

$$[Ni(NH_3)_2Cl_2]$$ → does not show any isomerism.

Hence, the correct answer is Option C.