The given square planar complex [Pt (Cl) (py) (NH3) (NH2OH)] consists of the central platinum atom bonded to four ligands - one chloride ion (Cl), one pyridine molecule (py), one ammonia molecule (NH3), and one hydroxylamine molecule (NH2OH).

To determine the number of geometric isomers, we need to consider the different ways the ligands can be arranged around the central platinum atom. There are two factors that contribute to the number of isomers:

1. The number of different ligands: In this complex, there are four different ligands - Cl, py, NH3, and NH2OH.

2. The number of ligands that can occupy different positions: In a square planar complex, the ligands can occupy either axial or equatorial positions. The axial positions are perpendicular to the plane of the molecule, while the equatorial positions lie within the plane.

Now, let's consider the different possible arrangements of ligands:

1. Axial Cl and equatorial py: In this arrangement, the Cl ligand occupies one axial position and the pyridine ligand occupies one equatorial position. The two remaining ligands, NH3 and NH2OH, can occupy the remaining axial and equatorial positions in any order. So, there are 2 possibilities for the arrangement of NH3 and NH2OH.

2. Axial py and equatorial Cl: In this arrangement, the pyridine ligand occupies one axial position and the Cl ligand occupies one equatorial position. The remaining two ligands, NH3 and NH2OH, can occupy the remaining axial and equatorial positions in any order. Again, there are 2 possibilities for the arrangement of NH3 and NH2OH.

3. Axial NH3 and equatorial py: In this arrangement, the NH3 ligand occupies one axial position and the pyridine ligand occupies one equatorial position. The remaining two ligands, Cl and NH2OH, can occupy the remaining axial and equatorial positions in any order. Once again, there are 2 possibilities for the arrangement of Cl and NH2OH.

Therefore, by considering all possible arrangements, we find that there are a total of 2 + 2 + 2 = 6 different geometric isomers for the given square planar complex [Pt (Cl) (py) (NH3) (NH2OH)].

Hence, the correct answer is option 'D' - 6.