Explain the enolisation process of 5,8-dihydroxynaphthalene-1,4-dione
There is no need for any catalyst apart from acid or base. Also, the $\ce{^{13}C}$ is just there to tell you that it is not the same molecule. Also, this is not a zwitterion, it is tautomerism since there is pi bond movement involved.
Don't deprotonate any carbons. They're all $\ce{sp^2}$ and won't take kindly to deprotonation. Just push electrons from one oxygen to the other through the conjugated system.
Try doing this one first. In acid, your general tautomerism mechanism would look like this:

Protonate the carbonyl group.
Push electrons from phenol oxygen.
Deprotonate the carbonyl formed on the other side.

In base, it would be:

Deprotonate the phenol.
Push electrons from phenoxide oxygen.
Reprotonate the phenoxide formed on the other side (strictly this shouldn't happen, but whatever).

If you can do this one, the mechanism for your original question just consists of the same thing repeated. You can &quot;fill in&quot; the parts of the molecule that I omitted to see what I mean.