Sphere on Incline: Kinetic Energy of a Rolling Sphere

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A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 3.9 m down a q = 35° incline. The sphere has a mass M = 4.6 kg and a radius R = 0.28 m.
https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys101/fall07/homework/08/03/3.gif

Suppose now that there is no frictional force between the sphere and the incline. Now, what is the translational kinetic energy of the sphere at the bottom of the incline?
 
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I figured out that the velocity is 5.6 m/s, but I don't know what the KE tran would be at the bottom without friction.
 
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The problem statement is incomplete, but I infer that you are being asked for the translational KE at the bottom for the two cases: the ball rolls down the incline, or it just slips down the incline (frictionless case).

If it just slips, then the KE final equals the PE initial, right? Why?

If it rolls, then some energy goes into the rolling motion (look up moment of inertia). What is the equation for the rotational energy of a sphere? What does that do to the final KE of the sphere at the bottom?