Assume $[\mathrm{Ni}(\mathrm{CN}){}_4]{}^{2-}$ is square planar because $\mathrm{CN}{}^{-}$ is a strong-field ligand. Its hybridisation and magnetic behaviour are:

$\mathrm{Ni}{}^{2+}$ is $3d^8$. The strong-field $\mathrm{CN}{}^{-}$ pairs the electrons in four $3d$ orbitals, freeing one $3d$ orbital. This $3d$, the $4s$ and two $4p$ orbitals give $ds{p}^2$ (square planar). No unpaired electrons remain, so the complex is diamagnetic.