(C) is $K_2[Ni(CN)_4]$ and $B$ is $K_2[NiCl_4]$.
$1.$ The $IUPAC$ name of $K_2[Ni(CN)_4]$ is Potassium tetracyanonickelate $(II)$ and $K_2[NiCl_4]$ is Potassium tetrachloronickelate $(II)$. Thus,option $(A)$ is correct.
$2.$ In $A$,$CN^-$ is a strong field ligand,so $Ni^{2+}$ $(3d^8)$ undergoes $dsp^2$ hybridization,making it diamagnetic. In $B$,$Cl^-$ is a weak field ligand,so $Ni^{2+}$ $(3d^8)$ undergoes $sp^3$ hybridization,leaving two unpaired electrons,making it paramagnetic. Thus,option $(C)$ is correct.
$3.$ As determined,$A$ has $dsp^2$ hybridization and $B$ has $sp^3$ hybridization. Thus,option $(A)$ is correct.
The sequence of answers is $A, C, A$.