1. Identify coordination number and likely geometry

For Pt complexes here, the ligands listed indicate coordination number $4$ or $6$:

• Coordination number $4$ for (a) and (c) $
• Coordination number $6$ for (b) and (d)

Geometrical isomerism depends on arrangement of ligands in the given geometry.

2. Check each complex

(a) $[Pt(NH_3)_3Cl]^+$

This is of type $[MA_3B]$ with coordination number $4$.

For a $4$-coordinate Pt complex, the relevant geometry is square planar. In square planar geometry, $[MA_3B]$ has only one arrangement because all positions become equivalent under symmetry.

So, no geometrical isomerism.

(b) $[Pt(NH_3)Cl_5]^-$

This is of type $[MAB_5]$ with coordination number $6$ (octahedral).

In an octahedral complex of type $[MAB_5]$, there is only one possible arrangement since the single $A$ ligand can be placed at any position and all such positions are equivalent.

So, no geometrical isomerism.

(c) $[Pt(NH_3)_2Cl(NO_2)]$

This is of type $[MA_2BC]$ with coordination number $4$.

For square planar complexes of type $[MA_2BC]$, geometrical isomerism is possible because the two identical ligands $A$ can be:

• cis to each other, or
• trans to each other

Hence, this complex can show geometrical isomerism.

(d) $[Pt(NH_3)_4ClBr]^{2+}$

This is of type $[MA_4BC]$ with coordination number $6$ (octahedral).

In octahedral complexes of type $[MA_4BC]$, the two different ligands $B$ and $C$ can be:

• cis to each other, or
• trans to each other

Hence, this complex can show geometrical isomerism.

3. Conclusion

Complexes showing geometrical isomerism are:

So the correct option is:

4. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They match.