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.
The electron states in the band gap are localised, whereas the states which contribute to the bands are not.
The electron states in a solid are not simple, there's a lot of non-trivial quantum mechanics going on. The electron states in a free atom are localised around the atom - the electrons in those states can't leave the atom without a lot of energy, so they can't conduct anything. When you pack lots of atoms together, the surrounding electron states overlap and mix. Which states mix with each other is dictated by the energy: similar energies means more mixing. You end up with a new set of states which extend over the whole block of material. If the material has a periodic lattice, these electron states group together into bands.
Every state in a band has some velocity (called a Fermi velocity) associated with it, and an electron in that state can be thought of as moving through the material with that velocity. The Fermi velocity of electrons in the conduction band is very large, but because the electrons are all going in different directions, there is no net current. An applied electric field moves some electrons from states which were going in one direction, to states going in the other. In a metal, one of the bands is part full, so there are plenty of nearby states to move electrons into. In a semiconductor, there is a gap between a full band and an empty one so it's much harder to push electrons into the higher band.
When dopants are added, they don't form a nice periodic lattice and they are much more spread out than the silicon atoms that host them. This means that the electron states around the dopant can't mix with states from other dopants to form a band. Since the energy levels of the dopant states are different from the silicon states, they don't mix (much) with them either. Instead, the electron states are localised around the dopant, much like the states around the free atom. An electron in that state can't conduct in the way one in a band can. It either has to jump up into the band, or jump to another nearby dopant. The former happens in semiconductors, the later is known as incoherent transport, and appears in some other materials.
I'm not sure how well I've explained this, but if you don't get a clear answer here, you could try the physics stack exchange. This definitely feels more like condensed matter physics than electrical engineering!
Best Answer
You are on the right lines with thinking about the electrons, but need to consider one more thing:
The energy band diagram of a p-n-junction at equilibrium is dependent on the Fermi Level for each material. The Fermi level is the theoretical energy level at which the probability of finding an electron is 50%. As a p-type material has an excess number of holes, i.e. fewer electrons, the Fermi level lies closer to the valence band. Then for an n-type material that has electrons occupying the conduction band, the energy at which the likelihood of finding an electron is 50% is therefore increased.
As at equilibrium, the Fermi levels must match in a p-n-junction, this leads to the p-type conduction band existing at a greater energy relative to the n-type conduction band.