Why does a neutral atom of beryllium not have any electrons in a p orbital?
We need to first consider n, the quantum number for the shell, and then l (~the classical angular momentum of the electron), which tells us the shape of the electron density. The first shell (n=1) contains two s electrons with paired spins. That shell is closed, and you begin over again with n=2: the possibilities for n=2 are l=0 (s electron shape, spherical) and l=1 (p electron shape, lobed). Then, since a lobed shape can point in different directions, when l=1, ml can equal +1, 0 or -1 , defining px, py, pz electron shapes, each pointing in one of three orthogonal directions.

So the filling process puts the first two (lowest energy) electrons in the 1s level (when full, it is 1s2); then the next level is 2s (filling to 2s2), then 2p. Since we have run out of electrons, the 2p levels do not get filled, so we don't have to consider 2px, 2py or 2pz designations.

Now, for boron, there will be a p electron. Whether you call it px, py or pz will depend on how you define the directions in a molecule. In an individual free boron atom, 2px, 2py and 2pz all have the same energy and are degenerate.

This is all theoretical. If you ever heated up beryllium so that there were free atoms, they would be so energetic that quantum levels up to 10 or 100 would be occupied, and there would be beryllium ions floating around too!