@aspodkfpo I answered that JEE question on stackexchange before, I'll try and copy/paste the question.
Interesting question, though I think that the questions could have been asked in a better way.
First, let met explain the angular velocity with respect to the ##z##-axis. If you look from atop, then you'd see something like this.
The smaller circle represents the projection of the position of the centre of mass of the disk onto the ##xy##-plane, while the bigger circle represents the actual path taken by the point of the disk which is in contact with the aforementioned plane. (I have only drawn the little circle, but it doesn't matter since they both roll with the same angular velocity).
Call the angle that is described on the ##xy##-plane ##\varphi##, then the angular displacement is given by ##\Delta s=\sqrt{l^2+a^2}\Delta\varphi##.
The crux of this problem lies in recognising that this angular displacement is equal to the angular displacement inside the disk! i.e ##\Delta s=a\Delta\phi## where I represent with ##\phi## the angle described as the disk rotates about its center of mass.
$$\begin{align*}
\sqrt{l^2+a^2}\Delta\varphi=a\Delta\phi
&\implies\Delta\varphi=\frac{a}{\sqrt{l^2+a^2}}\Delta\phi\\
&\implies\Delta\varphi=\frac{a}{\sqrt{25a^2}}\Delta\phi\\
&\implies\Delta\varphi=\frac{\omega t}{5}\\
&\implies\dot\varphi=\frac{\omega}{5}
\end{align*}$$
Now I'll explain the angular velocity about the origin. By this they meant you to imagine that the configuration is perfectly straight.
Do you see the line I have drawn? You need to imagine that it is as if the disks are rolling on that plane (we see it as a line from this point of view).
##\varphi'## is the angle of rotation on the imaginary plane.
The angular displacement is (the same arguments as before holds, but in this case the smaller disk doesn't touch the ground, so we operate on the bigger one)$$\Delta s=2l\Delta\varphi'=2a\Delta\phi\implies\dot\varphi'=\frac{a\omega}{l}=\frac{\omega}{\sqrt{24}}$$
To go from this back to first case, then if you wish to project##^*## ##2l## onto the ##xy##-plane you will do ##\left[2l/(\cos\theta)\right]\Delta\varphi=2a\Delta\phi##, hence ##\omega_z=\dot\varphi'\cos\theta##, which is why projecting the angular velocity pseudo-vector about the point ##O## in the imaginary plane onto the ##z##-axis works.
Possible confusion with regards to (##^*##):
Suppose that we have two referentials, ##(x,y,z)## and ##(x',y',z')##, which are on top of each other, i.e the same. Imagine that there is a point particle rotating about the origin in the ##x'y'##-plane such that its direction vector is ##\vec u=r\cos\phi\,\hat x'+r\sin\phi\,\hat y'##.
Now let us say that ##(x',y',z')## will rotate about the ##x##-axis such that ##\angle(\hat z,\hat z')=\angle(\hat y,\hat y')=\theta##, while ##\hat x=\hat x'##, then we get that ##\hat y'=\cos\theta\,\hat y+\sin\theta\,\hat z##, hence ##\vec u =r\cos\phi\,\hat x+r\sin\phi\cos\theta\,\hat y+r\sin\phi\sin\theta\,\hat z##.
If we project ##\vec u## onto the ##xy##-plane, we get ##\vec v=r\cos\phi\,\hat x+r\sin\phi\cos\theta\,\hat y## which definitely doesn't describe a circle!
However, this is not what we have in this situation, the "projection" here is in the process of bringing the whole apparatus from the imaginary position to what we see on the problem statement's image, thus only changing the length of the radius of rotation, but keeping the whole system moving in a circle.
Thus the projection is taking the new radius of rotation of the point of contact of either of the disks with the ground, which is painting a circle!
Conclusion: (I might be incorrect here, some time have passed since I have thought about this)
If you have a mobile point (here we considered the point with which one of the disks, doesn't matter which one, touches the floor) naturally rotating about a point ##O## with a radius of rotation ##r##, then, if it is brought to another level by rotation such that it still rotates in a circular fashion but about an axis that makes an angle ##\theta## with the original one, we can find that its angular velocity about this axis is equal to its angular velocity about the original axis scaled by ##\cos\theta##.
Addendum about the question on the point P:
The op there asked this:
b) I still don't quite realize why only chosing Origin led to the correct answer. For instance , Check out the point P i mention in the question. Why can't we take the cosin e of the angular velocity vector relative to point P?
Let's consider the point ##P##, and let's also place ourselves on the plane that contains ##P## and that is parallel to the ##xy##-plane in order to find an expression of our angle.
You can see that that ##z## height corresponds with the ##z## height of the center of mass of the little disk, hence our rotation radius is ##l\cosθ## and thus ##Δs=l\cosθΔφ##. We need to find a way to relate this to ##Δϕ## as I have done.
The point of contact is painting an angular dispalcement of ##Δs_1=aΔϕ##, while the center of mass in painting an angular displacement ##Δs_2=l\cosθΔφ##, but are both part of the same circle.
In general, two points of a rotating radius in a circle have this relation $$\frac{r_1\Delta\phi=\Delta s_1}{r_2\Delta\phi=\Delta s_2}=\frac{r_1}{r_2}$$ thus $$\frac{l\cos\theta\Delta\varphi}{a\Delta\phi}=\frac{l\cos\theta}{\sqrt{l^2+a^2}}$$ which means that $$\Delta\varphi=\frac{a\Delta\phi}{5a}=\frac{wt}{5}\implies\dot\varphi=\frac{\omega}{5}$$
. For instance, the direction of the angular velocity vector coincides with the instantaneous axis of rotation.
