Using the parallel axis theorem, a disk pendulum that is allowed to rotate around a fixed point with a distance $l$ from it's center, has total rotational kinetic energy $$E= \frac{1}{2}\omega^2(I_{cm}+ml^2)=\frac{1}{2}I_{cm}\omega^2 +\frac{1}{2}m\omega^2l^2$$ and since we can write the angular velocity in terms of the tangential velocity, i.e., $$\omega=\frac{v}{l}$$ where $v$ is the linear (tangential) velocity of the center of mass of the disk, then the above equation for the kinetic energy can simply be written $$E= \frac{1}{2}I_{cm}\omega^2 +\frac{1}{2}mv^2$$ and the disk has a moment of inertia $$I_{cm}=\frac{mR^2}{2}$$ which is the moment of inertia of thin disk (similar to a cylinder of small height) of radius $R$.
This means we can write its total energy as $$T=\frac{1}{2}I_{cm}\omega^2 +\frac{1}{2}mv^2+mgh$$
So in your above question, option (a) is the correct answer since option (b) ignores this second kinetic energy term altogether. It's also important to note that the second term in this expression has the velocity of the center of mass, which again can be expressed in terms of the angular velocity that the disc has around the fixed point i.e., $$KE_{p}=\frac{1}{2}m\omega^2l^2$$
and to note that the first term is the kinetic energy the disc has about its own axis. i.e., $$KE_{ax}=\frac{1}{2}I_{cm}\omega^2$$