Applying conditions of equilibrium to the stone

$\sum Fx= - N2+N1\cos75+ µN1\cos15 =0 \text {_______}(i) \\ \sum FY= -2000 - µN2+N1\sin75 - µN1\sin15 = 0 \text{ _____}(ii)$

Solving 2 equations, we get

$N1=2576.83 N \space \space \space \space N2=1289.19N $

Now, Applying conditions of equilibrium to the wedge

$\sum Fx= - N1\cos75 - µN1\cos15+P - µN =0 \text{_______}(i) \\ \sum FY= N -N1\sin75 + µN1\sin15 = 0\text{_____}(ii) $

Solving eqns (i) & (ii)

$N=2322.293 N$ & $P=1869.76 N $

The necessary force is $P=1869.76 N $