In Wilkinson's catalyst, the hybridization of central metal ion and its shape are respectively:

We begin by recalling what Wilkinson’s catalyst is. The commonly used notation is $$\text{RhCl(PPh}_3\text{)}_3$$, which means the central metal ion is $$\text{Rh}^{+1}$$ (rhodium in the +1 oxidation state) surrounded by one chloride ion and three triphenyl-phosphine ligands.

In order to decide the hybridization and the molecular shape, we count the total number of electron pairs (or equivalently, the total number of σ-bonds) around the metal centre. Each ligand donates a lone pair for a σ-bond.

We have:

$$\text{Number of ligands} = 1\;(\text{Cl}^-) + 3\;(\text{PPh}_3) = 4$$

So, there are four σ-bonds surrounding the Rh(I) centre.

The next step is to look at the preferred geometry for a transition-metal species that has four σ-bonds and a total valence electron count of 16. A 16-electron, d8 metal ion almost always adopts a square-planar arrangement rather than a tetrahedral one. This is seen in familiar complexes such as $$\text{[Ni(CN)}_4]^{2-}$$, $$\text{[PdCl}_4]^{2-}$$, and $$\text{[Pt(NH}_3)_4]^{2+}$$.

For a square-planar geometry, the hybrid orbitals used by the metal must be four coplanar orbitals obtained from one $$d$$, one $$s$$ and two $$p$$ orbitals. Symbolically, this is written as

$$\text{Hybridization} = dsp^2$$

Therefore,

$$\text{Hybridization of Rh} = dsp^2$$

$$\text{Shape} = \text{square planar}$$

Comparing these findings with the options given in the question:

Option A ⇒ $$sp^3d$$, trigonal bipyramidal   (incorrect)

Option B ⇒ $$d^2sp^3$$, octahedral   (incorrect)

Option C ⇒ $$dsp^2$$, square planar   (matches our result)

Option D ⇒ $$sp^3$$, tetrahedral   (incorrect)

Hence, the correct answer is Option C.