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Observe this case. We can make two choices regarding the axis of rotation. If we choose an axis of rotation that passes through the center of mass $G$ then the only force that would cause the ball to rotate is the friction force and some force $F$ that's acting on the center of mass would just provide translational motion.

However, if we choose the point $P$ as the axis of rotation then it is the opposite. In that case $F$ causes rotation and friction provides translational motion. However, I know that without friction no rolling is possible (the ball would just slide) but how is that possible if the force of friction contributes in no way to rotating while analyzing the instantaneous axis of rotation?

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I have modified your diagram a little

enter image description here

The torque does depend on the frictional force $f$ even when calculated about the axis through $P$ because the applied horizontal force $F$ and the frictional force are related.

Assuming the no slipping condition $a_{\rm G} = r \alpha$

The horizontal translational equation of motion for an applied force is given by $F-f = Ma_{\rm G}$ where $M$ is the mass of the cylinder and this equation holds irrespective of the axis that is considered for rotation.

For rotational motion about the centre of mass $fr=I_{\rm G}\alpha$.

Using the point of contact as the axis you get $Fr = I_{\rm P} \alpha$

However note that $F$ and $f$ are linked $F = f + Ma_{\rm G} = f+Mr\alpha$

So $Fr = (f+Mr\alpha)r = fr + Mr^2\alpha = I_{\rm P}\alpha = (I_{\rm G} +Mr^2)\alpha \Rightarrow fr = I_{\rm G}\alpha$

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