In the case of [Ni(CO)4] the Ni possess the electronic configuration of 3d84s2 but due to the presence of strong field ligand like CO the electrons in 4s orbitals gets paired up with unpaired electrons in 3d orbital and hence forms a outer orbital complex of sp3 i.e tetrahedral. In the case of [PdCl4]2− the 4d orbital of Pd2+ is diffused and therefore forms strong overlap with the orbitals of Cl and this leads to a larger splitting in the d orbitals and hence it forms square planar complex. Therefore the correct option is D.