Solution

When turning on the engine with a thrust force $F$ for a time $\Delta t$, the ship's momentum will be:
$$\vec{p}_1 = \vec{p} + \Delta\vec{p} \quad (1)$$
where
$$\Delta\vec{p} = \vec{F} \cdot \Delta t \quad (2)$$

Let's choose the direction of the x-axis and project equation (2):
$$\Delta p = F \cdot \Delta t \quad (3)$$

On the other hand, from equation (1) in projection onto the x-axis:
$$\Delta p = 2p \cdot \sin\frac{\alpha}{2} \quad (4)$$

Equating the right sides of equations (3) and (4):
$$F \cdot \Delta t = 2p \cdot \sin\frac{\alpha}{2}$$

From where the required engine burn time is:
$$\Delta t = \frac{2p \cdot \sin\frac{\alpha}{2}}{F}$$

The axis of the engine must be oriented to the original direction at an angle:
$$\beta = \frac{\pi}{2} + \frac{\alpha}{2}$$

Alternative solution:

It is necessary to add momentum:
$$\Delta p = p\sqrt{2 - 2\cos\alpha} = 2p \cdot \sin\left(\frac{\alpha}{2}\right)$$
Then the time:
$$\Delta t = \frac{\Delta p}{F} = \frac{2p}{F} \cdot \sin\left(\frac{\alpha}{2}\right)$$

The angle between the initial velocity and the force $F$ will be equal to the base angle of an isosceles triangle with the vertex angle $\alpha$:
$$\frac{1}{2}(\pi - \alpha)$$
The engine is oriented in the opposite direction to the force, meaning:
$$\beta = \pi - \frac{1}{2}(\pi - \alpha) = \frac{1}{2}(\pi + \alpha)$$