Calculate the speed of the 1.5 kg trolley at point P using the principle of conservation of mechanical energy.

Given:
Mass of trolley 1, $m_1 = 1.5$ kg
Initial height at A, $h_A = 2.0$ m
Initial velocity at A, $v_A = 0$ m/s (held stationary)
Height at P, $h_P = 0$ m (bottom of the incline)
Acceleration due to gravity, $g = 9.8$ m/s$^2$

According to the principle of conservation of mechanical energy:

Substitute the known values:

The speed of the 1.5 kg trolley at point P is $\boxed{\text{6.26 m/s}}$.

In an isolated system, the total linear momentum is conserved.

This is an inelastic collision where the trolleys stick together.
Given:
Mass of trolley 1, $m_1 = 1.5$ kg
Velocity of trolley 1 before collision, $v_{1i} = v_P = 6.26$ m/s (from Step 1)
Mass of trolley 2, $m_2 = 2$ kg
Velocity of trolley 2 before collision, $v_{2i} = 0$ m/s (stationary)
Combined mass after collision, $M = m_1 + m_2 = 1.5 \text{ kg} + 2 \text{ kg} = 3.5$ kg
Let $v_f$ be the final velocity of the combined trolleys.

According to the principle of conservation of linear momentum:

Substitute the known values:

The speed of the combined trolleys immediately after the collision is $\boxed{\text{2.68 m/s}}$.

After the collision, the combined trolleys move with constant velocity.
Given:
Velocity of combined trolleys, $v_f = 2.68$ m/s (from Step 3)
Time, $t = 3$ s

Distance travelled, $d = v_f \times t$

The distance travelled by the combined trolleys in 3 s after the collision is $\boxed{\text{8.04 m}}$.