Among $\left|Ni{\left(CO\right)}_4\right|,{\left|Ni{\left(CN\right)}_4\right|}^{2-},{\left|NiC{l}_4\right|}^{2-}$ species, the hybridisation state at Ni atom are respectively
]Atomic number of Ni = 28]

In $Ni{\left(CO\right)}_4$, nickel is $s{p}^3$ hybridised because in it oxidation state of Ni is zero. So, configuration of
${}_{28}Ni=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^8,4s^2$

(CO is a strong held ligand, hence pairing of electrons will occur)
In ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$, nickel is present as $N{i}^{2+}$, so its configuration $=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^8$

$C{N}^{-}$ is strong field ligand, hence it makes $N{i}^{2+}$ electrons to be paired up.
In ${\left[NiC{l}_4\right]}^{2-}$, nickel is present as $N{i}^{2+}$, so its configuration $=1s^2,2s^{2}2p^6,3s^{2}3p^6,3d^8$

$C{l}^{-}$ is weak field ligand, hence in $N{i}^{2+}$ electrons are not paired.