The geometry and magnetic behavior of the complex $\left[\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\right]$ are

Let oxidation state of $Ni$ be $: \text{'}x\text{'}.$
Oxidation state of $Ni$ in $\left[\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\right]$ can be calculated as:
$\mathrm{x} + 4\left(0\right) = 0,$ giving $\mathrm{x} = 0.$
Therefore, oxidation state of $Ni$ in $\text{[Ni}{\text{(CO)}}_4]$ is $0,$ giving its configuration to be $3{\mathrm{d}}^{8}4{\mathrm{s}}^2.$
CO pairs unpaired $3\mathrm{d}$ electrons because it is a strong field ligand. Additionally, it leads to a transfer of the 4S electrons into the 3d orbital, resulting in ${\mathrm{sp}}^3$ hybridization and the formation of the tetrahedral structure.
$\text{[Ni}{\text{(CO)}}_4]$ is diamagnetic since there aren't any unpaired electrons in this situation.
Hence, the correct option is option (B) tetrahedral geometry and diamagnetic.