The correct combination is:

First, we recall the ground-state electronic configuration of the free nickel atom. Nickel has atomic number 28, so in the isolated state

$$\text{Ni} : [\text{Ar}]\,3d^{8}\,4s^{2}.$$

When nickel forms complexes, the 4s electrons are lost first, and then, if necessary, 3d electrons are removed. The charge on the metal ion inside each complex therefore decides the d-electron count that must be used while discussing crystal-field splitting, hybridisation, geometry and magnetic behaviour.

We shall now analyse every complex appearing in the options one by one.

1. For $$[\text{NiCl}_{4}]^{2-}$$ we first work out the oxidation state of nickel.

Since each chloride carries a charge $$-1$$, the overall charge balance gives

$$x + 4(-1) = -2 \Longrightarrow x = +2.$$

Hence inside $$[\text{NiCl}_{4}]^{2-}$$ the metal centre is $$\text{Ni}^{2+}$$ and possesses

$$d^{8}\; \text{configuration (because }28-2 = 26\text{ and }26-18 = 8\text{ electrons in the d subshell).}$$

Chloride $$\left(\text{Cl}^{-}\right)$$ is a weak-field ligand. Crystal-field theory tells us that weak-field ligands produce a small splitting $$\Delta_{\text{tet}}$$ in a tetrahedral environment, so electrons prefer to remain unpaired rather than pair up. For a $$d^{8}$$ ion in a tetrahedral field the electronic distribution is

$$e\;( \uparrow \uparrow )^{4}\;t_{2}\;( \uparrow \uparrow )^{4},$$

leaving

$$2\;\text{unpaired electrons}.$$

Because no pairing energy is paid, tetrahedral geometry is energetically favored over square-planar for weak-field ligands. Therefore

$$[\text{NiCl}_{4}]^{2-}$$ is tetrahedral and paramagnetic (2 unpaired $$e^{-}$$).

2. For $$[\text{Ni(CN)}_{4}]^{2-}$$, we again equate charges:

$$x + 4(-1) = -2 \Longrightarrow x = +2,$$

so the ion is also $$\text{Ni}^{2+} (d^{8}).$$ However, the ligand now is $$\text{CN}^{-},$$ a strong-field ligand according to the spectrochemical series. A strong-field ligand causes a large splitting $$\Delta_{\text{sp}}$$ in a square-planar arrangement, large enough to make pairing of electrons cheaper than occupying the higher orbitals.

For a $$d^{8}$$ ion in a square-planar field we use the energy order

$$d_{x^{2}-y^{2}} \gt d_{xy} \gt d_{z^{2}} \gt d_{xz}, d_{yz}$$

and fill electrons pairwise in the lower three levels, giving

$$d_{xz}^{2}\,d_{yz}^{2}\,d_{z^{2}}^{2}\,d_{xy}^{2}\, d_{x^{2}-y^{2}}^{0}.$$

This leaves

$$0\;\text{unpaired electrons},$$

so the complex is

$$\text{square-planar and diamagnetic.}$$

3. For $$[\text{Ni(CO)}_{4}]$$ we determine the oxidation state:

Carbonyl $$\text{CO}$$ is a neutral ligand, the complex has no overall charge, so

$$x + 4(0) = 0 \Longrightarrow x = 0.$$

The metal is therefore $$\text{Ni}^{0}$$ with configuration

$$[\text{Ar}]\,3d^{8}\,4s^{2}.$$

Before complexation, both 4s electrons and the two 3d electrons occupying different orbitals shift so that all ten d electrons pair within the five 3d orbitals:

$$3d^{10},\;4s^{0}.$$

Now the metal employs four equivalent $$sp^{3}$$ hybrid orbitals (originating from 4s and 4p) to accept lone-pair donation from four CO molecules. The $$d^{10}$$ set remains intact and low in energy. Because CO is a very strong-field ligand, this arrangement leads to

$$\text{tetrahedral geometry with 0 unpaired}$$ $$\text{electrons (diamagnetic).}$$

Having established the facts, we inspect the statements in each option:

Option A: $$[\text{NiCl}_{4}]^{2-}$$ is not square-planar and $$[\text{Ni(CN)}_{4}]^{2-}$$ is not paramagnetic. Both parts are incorrect.

Option B: $$[\text{Ni(CN)}_{4}]^{2-}$$ is not tetrahedral and $$[\text{Ni(CO)}_{4}]$$ is not paramagnetic. Both parts are wrong.

Option C: $$[\text{NiCl}_{4}]^{2-}$$ is indeed paramagnetic (2 unpaired electrons) and $$[\text{Ni(CO)}_{4}]$$ is indeed tetrahedral. Both statements are correct.

Option D: $$[\text{NiCl}_{4}]^{2-}$$ is not diamagnetic and $$[\text{Ni(CO)}_{4}]$$ is not square-planar. Both parts are incorrect.

Thus only Option C matches all the established electronic structures, geometries and magnetic behaviours. Hence, the correct answer is Option C.