1. Key idea: rotation about the corner of the step

When the wheel just starts to climb the step and continues without slipping, the instantaneous pivot is the corner $Q$ of the step. So we examine the torque of the applied force about $Q$.

The torque magnitude is

where $d_\perp$ is the perpendicular distance of the line of action of the force from point $Q$.

Thus, for each option, we only need to inspect the line of action of the applied force.

2. Geometry of the wheel with respect to the step

The wheel has radius $R$ and the step has height $R$. Hence, when the wheel touches the corner $Q$, the center of the wheel is at a distance $R$ from $Q$.

Since the center is also at height $R$ above the ground and $Q$ is also at height $R$, the center lies horizontally level with $Q$. So the center $O$ is horizontally to the left of $Q$ by distance $R$.

Therefore, during climbing, the wheel rotates about $Q$ and the center moves on a circle of radius $R$ centered at $Q$.

3. Option A: Force applied normal to the circumference at point $P$

A force applied normal to the circumference at any point of a circle acts along the radius through that point, i.e. along a line passing through the center $O$.

At point $P$, the normal force therefore acts along $OP$. From the figure/setup, this line of action passes through $Q$ while the wheel climbs. Hence the perpendicular distance from $Q$ to the line of action is zero.

So,

Therefore, Option A is correct.

4. Option B: Force applied tangentially at point $S$

If the force is applied tangentially at point $S$, then its line of action is tangent to the wheel at $S$. The torque about $Q$ is nonzero if this tangent does not pass through $Q$. So the statement "$\tau \neq 0$" is acceptable.

But we must also check whether the wheel can climb.

A tangential force at a suitable point can produce a moment about $Q$ and make the wheel rotate upward about $Q$. Therefore the claim that the wheel never climbs the step is false.

Hence the combined statement in option B is false.

Therefore, Option B is incorrect.

5. Option C: Force applied at point $P$ tangentially

Now the force is applied at point $P$ tangentially.

As the wheel rotates about $Q$, the tangent at $P$ changes its position. Therefore the perpendicular distance from $Q$ to the tangent at $P$ changes continuously.

Let the center be $O$ and suppose the wheel has rotated by angle $\theta$ about $Q$. For the tangent at $P$, the torque magnitude about $Q$ is

where $d_\perp$ is the distance from $Q$ to the tangent at $P$.

Since the tangent at $P$ gradually comes closer (in lever-arm sense) to $Q$ as the wheel climbs, this distance decreases continuously. Hence $\tau$ decreases continuously.

So, Option C is correct.

6. Option D: Force applied normal to the circumference at point $X$

A force applied normal to the circumference at point $X$ acts along radius $OX$. As the wheel climbs, the direction of $OX$ changes, so the line of action changes continuously with rotation.

Therefore the perpendicular distance from $Q$ to this line of action is not constant in general. Hence the torque about $Q$ is not constant.

So, Option D is incorrect.

7. Final selection

The correct options are:

8. Comparison with stored correct answer

Stored correct answer: $A, C$

Our derived answer matches the stored answer exactly.