Step 1: Idea: count possible arrangements
A compound shows geometrical isomerism only if we can draw two different spatial arrangements that cannot be turned into each other without breaking bonds. I will test every species by drawing the arrangements.
Step 2: Option 1
Place four O atoms at the corners of a tetrahedron around Mn. Swap any two corners and the picture looks the same. Only one arrangement exists. So there is no isomerism here.
Step 3: Option 2
Place the metal at the centre of an octahedron with six corners. Put two $NH_3$ on opposite corners (trans) or on neighbouring corners (cis). These are two different molecules, so isomerism is present.
Step 4: Option 3
Each oxalate spans two neighbouring corners, so it must be cis within itself. The two Cl atoms fill the remaining corners. They can be on the same side or on opposite sides, giving two forms.
Step 5: Option 4
The en ligands behave like oxalate. The two Cl atoms again give a cis form and a trans form, so isomerism is present.
Final Answer:
Three of the four species have two arrangements, and the tetrahedral $MnO_4^-$ has just one. So the answer is option 1.
\[ \boxed{\text{Option 1}} \]