Why is the positive charge not distributed in the entire N type region (as there are free electrons for conduction in the N type region)?
It actually is. The diagrams are shown that way because we're primarily interested in the electric field produced by that charge, which is strongest near the junction, since that's where it's nearest the corresponding negative charge.
Why are the charges fixed at junction despite the presence of holes and free electrons?
Once the diffusion has stabilized, there are no free carriers in the central region, which becomes insulating. This insulating barrier becomes the dielectric between the two conductive regions, effectively forming a capacitor.
How can a charge neutral substance have a potential?
To say that a PN junction has built-in potential isn't to say that the PN junction has a potential relative to ground or infinity etc.
How can there be an inherent potential in a doped semi-conductor if it
is charge neutral?
Charge has been separated within the PN junction and, thus, there is an electric field across the depletion region and an associated potential difference. A charged capacitor is neutral but there is a potential difference (voltage) across the dielectric. There are some similarities but...
However, I don't understand why we can't measure the voltage drop
across the PN block with a voltmeter.
As explained, for example, here, the built-in potential is not readily measured with, e.g., a voltmeter. In other words, this question has been asked here several times (which means you are not the only one perplexed by this - most are at first) and there are good answers already available.
Best Answer
It is not. The current is zero (because it is in open-circuit condition, no path for the flow.)