Step 1:

In [Cr(NH₃)₆]³⁺, Cr is in the +3 oxidation state i.e., 3$d^3$ configuration. Also, $N{H}_3$ is a weak field ligand that does not cause the pairing of the electrons in the 3d orbital. $C{r}^{3+}$

Therefore, it undergoes $d^{2}s{p}^3$ hybridization and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.

Step 2:

In [Ni(CN)₄]^2-, Ni oxidation is +2 and its electronic configuration is 3d⁸ configuration.

${\mathrm{Ni}}^{2+\; =\hspace{0.17em}}$

$C{N}^{-}$ is a strong field ligand. Hence, it paired the 3d electrons.  Then, $N{i}^{2+}$ undergoes $ds{p}^2$ hybridization.

As there are no unpaired electrons, it is diamagnetic in nature.