Kinetic energy of rigid body [closed]

A disc of mass mass $M$,radius $R$ is in state of pure rolling upon a horizontal plane as shown in figure. Find their kinetic energy of rigid body in the reference frame of P which is directly below the centre of disc, at a distance of $\frac{R}{4}$

K.E of B =Rotational Kinetic Energy+ Translational Kinetic Energy $$=\frac{1}{2}\left(\frac{MR^2}{2}+ \frac{MR^2}{16}\right){\omega}^2 + \frac{1}{2}MV^2$$ $$=\frac{9}{32}MV^2+\frac{1}{2}MV^2$$ But the answer says it is only $=\frac{9}{32}MV^2$.Why it only rotational and why translational kinetic energy is zero?

closed as off-topic by AccidentalFourierTransform, Jon Custer, sammy gerbil, Chris♦, Kyle KanosJan 31 '18 at 10:55

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In the frame of reference of $P$ what do you see?
You see point $P$ not moving and the disc rotating at angular speed $\Omega$ about "fixed" point $P$.
• About $C$ the moment of inertia of the disc will be different. – Farcher Sep 18 '16 at 6:06