Answer
We can rank the forces according to the magnitude of the torque they create about point $P$:
$F_5 \gt F_4 \gt F_2 \gt F_1 \gt F_3$
Work Step by Step
We can write a general expression for the torque:
$\tau = r \times F$
$r$ is the displacement vector from the axis of rotation to the point where the force is applied
$F$ is the force vector
The torque increases as the magnitude of $r$ and $F$ increase.
The torque is a maximum when the angle between these two vectors is $90^{\circ}$ and the torque decreases to zero as the angle $\theta$ decreases to zero.
The angle between $r_3$ and $F_3$ is zero. Therefore, the torque due to $F_3$ is zero.
The angle between $r_1$ and $F_1$ is slightly more than zero. Therefore, the torque due to $F_1$ is slightly more than zero.
The angle between $r_2$ and $F_2$ is slightly more than the case with $F_1$. Therefore, the torque due to $F_2$ is slightly more than the torque due to $F_1$
If $F_4$ is extended to where it makes a $90^{\circ}$ with the displacement vector from point P, we can see that the torque due to $F_4$ is greater than the torque due to $F_2$
The torque due to $F_5$ is clearly greater than the torque due to $F_4$.
We can rank the forces according to the magnitude of the torque they create about point $P$:
$F_5 \gt F_4 \gt F_2 \gt F_1 \gt F_3$
