(N/A) In both complexes,$Ni$ is in the $+2$ oxidation state,which corresponds to a $d^8$ electronic configuration.
For $[Ni(CN)_4]^{2-}$:
$CN^-$ is a strong field ligand. It causes the pairing of the two unpaired $3d$ electrons. This results in $dsp^2$ hybridization,leading to a square planar geometry. Since all electrons are paired,the complex is diamagnetic.
For $[NiCl_4]^{2-}$:
$Cl^-$ is a weak field ligand. It does not cause the pairing of the two unpaired $3d$ electrons. Therefore,the complex undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry. Due to the presence of two unpaired electrons,the complex is paramagnetic.