From the FBD of 10 kN block on the inclined plane
Sum up forces normal to the incline
$N_2 = 10 \cos 30^\circ = 8.66 \, \text{ kN}$
Amount of friction
$f_2 = \mu N_2 = 0.20(8.66) = 1.732 \, \text{ kN}$
Sum up forces parallel to the incline
$T = f_2 + 10 \sin 30^\circ$
$T = 1.732 + 10 \sin 30^\circ$
$T = 6.732 \, \text{ kN}$
From the FBD of 30 kN block on the horizontal plane
Sum up vertical forces
$N_1 + P \sin a = 30$
$N_1 = 30 - P \sin \alpha$
Amount of friction
$f_1 = \mu N_1 = 0.20(30 - P \sin \alpha)$
$f_1 = 6 - 0.20P \sin \alpha$
Sum up horizontal forces
$P \cos \alpha = f_1 + T$
$P \cos \alpha = (6 - 0.20P \sin \alpha) + 6.732$
$P \cos \alpha + 0.20P \sin \alpha = 12.732$
$P (\cos \alpha + 0.20 \sin \alpha) = 12.732$
$P = \dfrac{12.732}{\cos \alpha + 0.20 \sin \alpha}$
To minimize P, differentiate then equate to zero
$\dfrac{dP}{d\alpha} = \dfrac{-12.732(-\sin \alpha + 0.20 \cos \alpha)}{(\cos \alpha + 0.20 \sin \alpha)^2} = 0$
$-\sin \alpha + 0.20 \cos \alpha = 0$
$\sin \alpha = 0.20 \cos \alpha$
$\tan \alpha = 0.20$
$\alpha = 11.31^\circ$
Thus,
$P_{min} = \dfrac{12.732}{\cos 11.31^\circ + 0.20 \sin 11.31^\circ}$
$P_{min} = 12.5 \, \text{ kN}$ answer
