1. Identify the complex type

The complex is $[\text{Pt}(\text{Cl})(\text{py})(\text{NH}_3)(\text{NH}_2\text{OH})]^+$

Pt(II) complexes are typically square planar.

Here, all four ligands are different monodentate ligands:

• $\text{Cl}$
• $\text{py}$
• $\text{NH}_3$
• $\text{NH}_2\text{OH}$

So this is of the type: $[Mabcd]$ for a square planar complex.

2. Geometrical isomerism in square planar $[Mabcd]$

In a square planar complex, geometrical isomerism depends on which ligands are trans to each other.

Since there are 4 different ligands, a geometric isomer is determined by choosing one pair of ligands to be trans; the remaining two automatically become trans to each other.

The number of distinct ways to partition 4 different ligands into 2 trans pairs is: $\frac{1}{2}\binom{4}{2} = 3$

These are:

• $(\text{Cl} \text{ trans py})$ and $(\text{NH}_3 \text{ trans NH}_2\text{OH})$
• $(\text{Cl} \text{ trans NH}_3)$ and $(\text{py} \text{ trans NH}_2\text{OH})$
• $(\text{Cl} \text{ trans NH}_2\text{OH})$ and $(\text{py} \text{ trans NH}_3)$

Thus, there are 3 distinct geometric isomers.

3. Check options

• A: 3 ✓
• B: 4 ✗
• C: 6 ✗
• D: 2 ✗

4. Final answer

The number of geometric isomers is: $\boxed{3}$