Method to analyse double rotation of bodies

This problem hails from the 'JEE (Advanced) 2021' -

A thin rod of mass $M$ and length $a$ is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass $M$ and of radius $\frac a 4$ is pivoted on this rod with its center at a distance $\frac a 4$ from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4\Omega$. The total angular momentum of the system about the point O is $\left (\frac {Ma^2\Omega}{48}\right)n$. The value of $n$ is …

The solutions provided by a hoard of resources involve calculating the rotational mass of the rod about its end points, the disc about its center, multiplying them with their corresponding angular velocities and then adding the '$mvr term: $ L = \frac{Ma^2}3\Omega+\frac 1 2 {M(\frac a 4)^2}(4\Omega)+M\left(\frac{3a}4\right)\left(\frac{3a}4\Omega\right) $ Thereby yielding the correct answer.

However, I find a little confusing since last term (to me) appears as if we have concentrated the mass of the entire disc about its center of mass and then calculated it's angular momentum.

Per my understanding, last term should be the expression of the disc's rotational mass about 'O' multiplied by $\Omega$ (it's angular velocity of orbit): $ \left(\frac 1 2 M\left (\frac a 4\right)^2+M\left(\frac{3a}4\right)^2\right)\Omega $

Why does this reasoning fall short?