How can the overvoltage prodection inhibit the voltage to let the voltage doesn't reach the destrutive level?
Consider the following circuit. It might not be even remotely close to the same circuit, but it suffices to explain the OVP part.







simulate this circuit &ndash; Schematic created using CircuitLab

When Movp is cutoff (normal operation - no overvoltage), it is equivalent to this circuit when Rovp is very high, to the point that it could be considered infinite for any practical analysis. The voltage Vbuck is therefore given as a ratio of the voltage source, nominally Rldo/(Rldo+Rbuck), and power consumed by LDO is $$\frac{Vbuck^2}{Rldo}$$

When Movp is activated (overvoltage detected), then Rovp in our circuit goes down. If it essentially becomes a short circuit, Vbuck goes to 0, ending the overvoltage. If the channel resistance Rovp is still to be considered, then 
the total power consumed by LDO and OVP will become $$\frac{Vbuck^2}{Rldo} + \frac{Vbuck^2}{Rovp}$$

The circuit designer then expects this total output power to be too much for the overvoltage source to handle, killing the overvoltage.