Nice problem!
Note, that the angular velocity of the sphere is NOT $V/r$. There is a component of the angular velocity directed along the vertical axis.
Imagine another, simpler problem. It's almost the same, but the sphere is not rolling on a table. It is sliding along it.
It's kinetic energy would not be $V^2/R$. Because the sphere is actually rotating - you will see it if look at the sphere from above!
UPDATE.
Formula $E=mv^2 + I\omega^2/2$ (where $v$ is the velocity of center of mass) is correct. In this particular problem it is very easy to make a mistake calculating the angular velocity $\omega$ and so get incorrect final answer.
Looks like the axle of rotation of the sphere at any moment is $OP$ - the line which goes via $O$ and center of the sphere $P$. But this is not actually so!
In the frame of reference which does not rotate but is moving with the same velocity $\vec{v}$ as the center of mass, velocity of any point of the body is $\vec{v}(\vec{r}) = [\vec{w}*\vec{r}$] , where $\vec{r}$ is a vector from the center of mass to our point of the body. For all the points along the axle of rotation this velocity is zero.
In the original frame of reference all these points should have the same velocity (same as the velocity of the center of mass).
But clearly the velocities of different points of sphere located along the axle $OP$ are different - further from $O$, bigger the speed. So, $OP$ is not the axle of rotation of the sphere!
Well, if you get into the frame of reference which rotates around point $O$ with angular velocity $W=V/R$ the speed of every point along the $OP$ would be zero. This would be the axle of rotation of sphere, and in this frame of reference the angular velocity of the sphere would indeed be $V/r$. And to find the angular velocity in the original frame of reference you need to add up $\vec{w}$ and $\vec{W}$ - but remember that both of them are vectors and you should add them as vectors!