An object spinning about a point not the center of mass will have translational kinetic energy since it has momentum.
To find the proportion of kinetic energy due to translation and the proportion due to rotation you must express the kinetic energy at the center of mass, and then the $\tfrac{1}{2} m v_{\rm CM}^2$ part is the translating energy and the $\tfrac{1}{2} I_{\rm CM} \omega^2$ is the rotational energy.
Consider a rod pinned on one end. It has at the center of mass $I_{\rm CM} = \tfrac{m}{12} \ell^2$ and $v_{\rm CM} = \omega \tfrac{\ell}{2}$
$$ \begin{aligned}
K & = \tfrac{1}{2} m v_{\rm CM}^2 + \tfrac{1}{2} I_{\rm CM} \omega \\
& = \underbrace{ \tfrac{1}{8} m \ell^2 \omega^2 }_\text{trans} + \underbrace{ \tfrac{1}{24} m \ell^2 \omega^2 }_\text{rot} \\
& = \underbrace{ \tfrac{1}{6} m \ell^2 \omega^2 }_\text{total}
\end{aligned}$$
But take the same situation and calculate the kinetic energy at the pivot A with $I_{\rm A} = \tfrac{m}{3} \ell^2$ and $v_{\rm A} = 0$
$$ \begin{aligned}
K & = \tfrac{1}{2} m v_{\rm A}^2 + \tfrac{1}{2} I_{\rm A} \omega \\
& = 0 + \tfrac{1}{2} ( \tfrac{1}{3} m \ell^2) \omega^2 \\
& = \underbrace{ \tfrac{1}{6} m \ell^2 \omega^2 }_\text{total}
\end{aligned}$$
Same total kinetic energy, but the proportion of translating and rotating isn't evident at all.
In the case where the body is purely translating with $v_{\rm CM} \neq 0$ and $\omega =0$ then the kinetic energy calculation is
$$ \require{cancel} K = \frac{1}{2} m v_{\rm CM} + \cancel{ \frac{1}{2} I_{\rm CM} \omega} $$
