1. Check the geometry and possibility of chirality for each complex

Optical isomerism in coordination compounds arises when the complex is non-superimposable on its mirror image and has no plane/center of symmetry.

A useful rule:

• Octahedral complexes of type $[M(AA)_3]$ where $AA$ is a bidentate ligand often show optical isomerism.
• Tetrahedral complexes generally do not show optical isomerism unless all four groups are different.

2. Option A: $[Zn(en)(NH_3)_2]^{2+}$

• $en$ is bidentate, occupying 2 coordination sites.
• Two $NH_3$ ligands occupy 2 more sites.
• Total coordination number $=4$.
• $Zn^{2+}$ with coordination number 4 is generally tetrahedral.

Thus this is a tetrahedral complex with ligand set effectively of type $M(AA)B_2$. Such a structure has symmetry and is not optically active.

So, A does not show optical isomerism.

3. Option B: $[Co(en)_3]^{3+}$

• Each $en$ is bidentate.
• Three $en$ ligands occupy all 6 coordination sites.
• So the complex is octahedral of type $[M(AA)_3]$.

This is the classic case of optical isomerism, giving:

• $\Delta$ form
• $\Lambda$ form

These are non-superimposable mirror images.

So, B shows optical isomerism.

4. Option C: $[Co(H_2O)_4en]^{3+}$

• One $en$ gives 2 sites, four $H_2O$ give 4 sites.
• Total coordination number $=6$, so octahedral.
• Type: $[M(AA)B_4]$.

In an octahedral complex with one symmetrical bidentate ligand and four identical monodentate ligands, the complex has symmetry elements and does not become chiral.

So, C does not show optical isomerism.

5. Option D: $[Zn(en)_2]^{2+}$

• Two $en$ ligands provide 4 donor atoms.
• Coordination number $=4$.
• For $Zn^{2+}$, this is generally tetrahedral.

A tetrahedral complex with two identical bidentate ligands does not produce a stable pair of optical isomers here.

So, D does not show optical isomerism.

6. Final conclusion

Only the octahedral tris-chelate complex $[Co(en)_3]^{3+}$ is optically active.

Therefore, the correct option is:

$\boxed{B}$