How can we able to explain the vapour pressure in between solid and liquid phase in case of freezing point?
Start by considering the vapour in equilibrium with the liquid. For the two to be in equilibrium they must have the same molar Gibbs free energy. As a rough approximation the Gibbs free energy of the gas depends most strongly on its pressure while the Gibbs free energy of a liquid depends most strongly on its temperature. So for the two to stay the same the pressure of the vapour has to rise as we increase the temperature of the liquid, and vice versa.
The point of this is that the vapour pressure is a measure of the Gibbs free energy of the liquid. It isn't a simple proportionality because the equations for the Gibbs free energies are complicated, but we the two to be related by some smooth curve.
Now we can apply the same argument to the vapour in equilibrium with the solid, so again we find the vapour pressure is a measure of the Gibbs free energy of the solid. This means that if the vapour pressure of the liquid and the vapour pressure of the solid are the same then the Gibbs free energies of the liquid and solid are the same. But if the the Gibbs free energies of the liquid and solid are the same that means the liquid and solid are in equilibrium i.e. this happens at the melting point of the solid.
So suppose we graph the vapour pressures of the liquid and solid on the same graph we will get something like this:
The solid lines are the real experimental data for water and ice. The dashed lines are a freehand extrapolation of the solid lines drawn to guide the eye.
The melting point is the point where the two curves intersect, which of course happens at $0$°C for water.
So this is how you relate the vapour pressure to the melting point for any material. Find the curves describing the vapour pressures of the liquid and solid, and the melting point will be where the two curves intersect.