Electronic – Doping concentration in anode region

I think, the answer is relatively simple. Do you know the working principle of a "Schottky diode", which is based on a semiconductor-metal junction? Now - what happens if you connect a voltmeter (or any other load) across the diode? You create two Schottky junctions which exactly compensate the diffusion voltage inside the pn diode. Thus, no voltage can be measured. With other words: You cannot use the diffusion voltage to drive any current through an external load.

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.