According to the valence bond theory the hybridization of central metal atom is dsp$$^2$$ for which one of the following compounds?

In valence bond theory, dsp$$^2$$ hybridisation leads to a square planar geometry and is characteristic of certain transition metal complexes, particularly those of Ni$$^{2+}$$ with strong field ligands.

NiCl$$_2$$·6H$$_2$$O (option 1): The complex ion is $$[\text{Ni(H}_2\text{O)}_6]^{2+}$$, which is octahedral. Water is a moderate to weak field ligand for Ni$$^{2+}$$, and the complex is octahedral with sp$$^3$$d$$^2$$ hybridisation. Not dsp$$^2$$.

K$$_2$$[Ni(CN)$$_4$$] (option 2): The complex ion is $$[\text{Ni(CN)}_4]^{2-}$$, where Ni is in $$+2$$ oxidation state (d$$^8$$ configuration). CN$$^-$$ is a very strong field ligand. For Ni$$^{2+}$$ (d$$^8$$) with 4 strong field ligands, the electrons pair up in the d orbitals, leaving one empty d orbital available for hybridisation. The hybridisation is dsp$$^2$$, giving a square planar geometry. This complex is diamagnetic and square planar.

[Ni(CO)$$_4$$] (option 3): CO is a strong field ligand, but here Ni is in the zero oxidation state (d$$^{10}$$ configuration). With a d$$^{10}$$ configuration and 4 ligands, the hybridisation is sp$$^3$$ (tetrahedral), not dsp$$^2$$.

Na$$_2$$[NiCl$$_4$$] (option 4): The complex ion is $$[\text{NiCl}_4]^{2-}$$, where Cl$$^-$$ is a weak field ligand. For Ni$$^{2+}$$ (d$$^8$$) with weak field ligands, the hybridisation is sp$$^3$$, giving a tetrahedral geometry with 2 unpaired electrons.

Therefore, the compound with dsp$$^2$$ hybridisation is K$$_2$$[Ni(CN)$$_4$$], which is option 2.