1. Identify the complex and metal ion

The complex is $NiCl_2\left[ P\big((C_2H_5)_2(C_6H_5)\big)\right]_2$ which is of the type $NiCl_2L_2$ where $L$ is a neutral phosphine ligand.

Nickel is in the $+2$ oxidation state: $Ni^{2+} : 3d^8$

So we need to determine which geometry of a $d^8$ $Ni^{2+}$ complex is paramagnetic and which is diamagnetic.

2. Possible 4-coordinate geometries for $Ni^{2+}(d^8)$

A 4-coordinate $Ni^{2+}$ complex commonly adopts either:

• tetrahedral, or
• square planar

These two geometries have different magnetic properties.

3. Magnetic behavior in tetrahedral geometry

For a tetrahedral field, the crystal field splitting is small. Thus for $d^8$, electrons remain unpaired.

Tetrahedral $Ni^{2+}(d^8)$ typically has 2 unpaired electrons, so it is: $\text{paramagnetic}$

4. Magnetic behavior in square planar geometry

For square planar $d^8$ complexes, the splitting pattern strongly favors electron pairing. All electrons become paired.

Thus square planar $Ni^{2+}(d^8)$ is: $\text{diamagnetic}$

5. Apply to the question

The compound shows temperature-dependent magnetic behavior, meaning it can exist in two forms:

• paramagnetic form $\Rightarrow$ tetrahedral
• diamagnetic form $\Rightarrow$ square planar

So, the coordination geometries of $Ni^{2+}$ in the paramagnetic and diamagnetic states, respectively, are: $\boxed{\text{tetrahedral and square planar}}$

6. Check options

• A: tetrahedral and tetrahedral $\rightarrow$ wrong
• B: square planar and square planar $\rightarrow$ wrong
• C: tetrahedral and square planar $\rightarrow$ correct
• D: square planar and tetrahedral $\rightarrow$ wrong

Therefore, the correct answer is: $\boxed{\text{C}}$