Explain on the basis of valence bond theory that tetracyanonickelate(II) ion with square planar structure is diamagnetic
@M.L has already provided the solution as per VBT, but that answer seems to have some flaws when it comes to CFT (looks like he has corrected the mistakes now). So, I would like to take some time to answer this question using CFT.

According to crystal field theory, the degeneracy of the orbitals of a same orbital is lost in presence of a ligand and the pattern in which they are distorted depends upon the nature of the ligand. The spectrochemical series helps us understand the nature of a ligand.
In general there are two types of ligands, strong field and weak field. Strong field ligands have a greater power to distort the degeneracy of the sub-orbitals whereas weak field have a weaker power of distortion.
As per the spectrochemical series, $\ce{CN-}$ is a strong field ligand and thus creates a greater distortion in the degeneracy of $\mathrm d$ orbitals.
The central metal ion here is the $\ce{Ni^2+}$ with its $\mathrm d^8$ electronic configuration.
A $\mathrm d^8$ system prefers a square planar approach of ligands if the ligands are strong field. In such a condition the the degeneracy is disrupted and the energy diagram of the $\mathrm d$ orbitals is as below:

Now when trying to fill this diagram with electrons we have to take into consideration that an electron in the $\mathrm d_{x^2-y^2}$ orbital can pair up with the electron in $\mathrm d_{xy}$ in case of a strong field ligand. This is because the extent of splitting done by a strong field ligand is so high that it can overcome the energy gained due to pairing of electrons.
Thus the energy diagram filled with electrons looks as following:

Now the ligands donate their lone pairs in the $3\mathrm d_{x^2-y^2}$, $4\mathrm s$, $4\mathrm p_x$ and $4\mathrm p_y$ sub-orbitals.
In reality, no hybridization takes place but if forced to say the hybridization of this complex, we can say $\mathrm{dsp}^2$ because of the orbitals involved in the $\sigma$ bonds between the metal and ligands.
As you can see the energy diagram does not contain any unpaired electron and thus the complex is diamagnetic.