Given: mass m = 12 kg, angle of inclination theta = 30^, coefficient of friction mu = (2)/(3), and g = 10 ms⁻².

Since the body would not slide down on its own ($W_{\parallel} < F_f$), the external force $P$ must be acting down the plane to initiate motion.
At the point of impending motion down the plane, the forces acting down the plane are balanced by the forces acting up the plane. Friction acts up the plane, opposing the downward motion.

Substitute the calculated values:

Solve for $P$:

The value of $P$ is $\boxed{(40\sqrt{3} - 60) \text{ N}}$.

For the body to move up the plane, the external force $P$ must be acting up the plane.
At the point of impending motion up the plane, the forces acting up the plane are balanced by the forces acting down the plane. Friction acts down the plane, opposing the upward motion.

Substitute the calculated values:

The value of $P$ is $\boxed{(60 + 40\sqrt{3}) \text{ N}}$.