I don't think your question can be answered without looking in detail at the engineering mechanics of the tyres, bearing friction and losses in any skid.
The centripetal force for an object is normal to the velocity vector, so does no work. The only work done when there is acceleration along the tangent vector $\hat{\vec{t}}$ to the path:
$${\rm d}_t \vec{v} = {\rm d}_t (v\, \hat{\vec{t}}) = \dot{v} \hat{\vec{t}} + \kappa\,v^2\,\hat{\vec{n}}$$
the second step following from the Frenet–Serret formula (here $\hat{\vec{n}}$ is the unit normal and $\kappa$ path's local curvature), so that instantaneous rate at which energy is input to / lost from the bicycle/rider system is:
$$p = m\, \left<\vec{v},{\rm d}_t \vec{v}\right> = m\, v\,\dot{v} = \frac{m}{2}\,{\rm d}_t(v^2)$$
(whence the kinetic energy formula, btw: you can see it's quite general). To my mind, the only power transfer if the speed is constant is the power dissipated irreversibly deforming the tires and in bearing friction, so this is not a trivial question. Probably the first mechanism will be the main one: to understand this, witness that there must always be either skid or local deformation of a tyre with a nonzero contact area, a proposition whose truth you can grasp by thinking about a nonzero length cylinder rolling on a curved path so that the velocity vector is tangent to the cylinder's axis of rotational symmetry. The relationship $v = \omega\,R$ where $R$ is the radius of curvature can hold only at one plane normal to the cylinder's axis, if the cylinder is rigid (so that $\omega$ is the same everywhere). Put it another way, if no part of the cylinder is skidding, and if $R$ is the radius of curvature of the path of the cylinder's centre of mass, then the angular speed of the plane in the cylinder a distance $\ell$ along the axis would need to fulfill $v(\ell) = \omega(\ell) (R + \ell)$ if each plane cross section stays underformed. This cannot happen indefinitely as the cylinder would undergo torsion and fail from torsional shear. So in practice, what happens in a nonskidding tyre is that the rubber on the inside edge gets deformed so that its angular speed is slower than the rest of the rubber on the same edge, then, when it leaves the ground, it flicks forward to "catch up" with the rest of the rubber. An analogous description holds for each tyre cross-section aside from the one fulfilling $v = \omega\,R$. Moreover, there is a cyclic torsion about the tyre's radius as the rubber contacts the ground, stays so fleetingly and then lets slip its grip. So the tyre at most of its points is in a constant stretch-shrink deformation cycle locally with period $2\pi/\omega$ and this is what dissipates most energy in turns and also wears out the tyres.