Summation of forces normal to the incline in the 200-lb block
$N_2 = 350 \cos 45^\circ + C \cos 75^\circ$
$N_2 = 247.49 + 0.2588C$
Amount of friction under the 200-lb block
$f_2 = \mu N_2 = 0.20(247.49 + 0.2588C)$
$f_2 = 49.498 + 0.0518C$
Summation of forces parallel to the incline in the 200-lb block
$f_2 + 350 \sin 45^\circ = C \sin 75^\circ$
$(49.498 + 0.0518C) + 350 \sin 45^\circ = C \sin 75^\circ$
$0.9141C = 296.985$
$C = 324.89 \, \text{ lb}$
Summation of all vertical forces acting in the 400-lb block
$N_1 = 550 + C \sin 30^\circ$
$N_1 = 550 + 324.89 \sin 30^\circ$
$N_1 = 712.45 \, \text{ lb}$
Amount of friction under the 400-lb block
$f_1 = \mu N_1 = 0.20(712.45)$
$f_1 = 142.50 \, \text{ lb}$
Solving for the required P by summing up horizontal forces in the 400-lb block
$P = f_1 + C \cos 30^\circ$
$P = 142.50 + 324.89 \cos 30^\circ$
$P = 423.85 \, \text{ lb}$ answer
