1. Identify the coordination geometry

All the given cobalt complexes are octahedral $6$-coordinate complexes.

2. Recall the condition for optical isomerism

A coordination compound shows optical isomerism if it is chiral, i.e. it has no plane of symmetry, no center of symmetry, and its mirror image is non-superimposable.

We now check each option.

3. Option A: $\text{cis}[Co(en)_2Cl_2]Cl$

The complex ion is $[Co(en)_2Cl_2]^+$.

• It is an octahedral complex of type $[M(en)_2a_2]$.
• Since $en$ is a bidentate ligand, it forms chelate rings.
• In the cis form, the arrangement of the two $en$ rings creates a helical, non-superimposable mirror image.
• Thus the cis form is chiral and shows optical isomerism.

So, Option A shows optical isomerism.

4. Option B: $\text{trans}[Co(en)_2Cl_2]Cl$

The complex ion is $[Co(en)_2Cl_2]^+$.

• In the trans form, the two identical $Cl^-$ ligands are opposite each other.
• This arrangement has symmetry elements (in particular, it is not chiral).
• Therefore it does not show optical isomerism.

So, Option B is not optically active.

5. Option C: $[Co(NH_3)_4Cl_2]Cl$

The complex ion is $[Co(NH_3)_4Cl_2]^+$, of type $[MA_4B_2]$.

• Such complexes show geometrical isomerism (cis/trans).
• But because all ligands are monodentate, these forms have symmetry and are not optically active.

So, Option C does not show optical isomerism.

6. Option D: $[Co(NH_3)_3Cl_3]$

This is of type $[MA_3B_3]$.

• It can show fac/mer geometrical isomerism.
• However, neither fac nor mer form is optically active in this case.

So, Option D does not show optical isomerism.

7. Final conclusion

Only $\text{cis}[Co(en)_2Cl_2]Cl$ shows optical isomerism.

Therefore, the correct answer is:

$\boxed{A}$

8. Comparison with stored correct answer

Stored correct answer: $A$

My derived answer: $A$

They agree.