P substrate has electrons?
What is conductivity? Part of the idea is the density of free electrons in the solid. Electrons are normally part of a neutral atom. But heat agitates the atoms and the vibrations can kick some of the electrons momentarily up into a conduction band, where they are easily affected by an electric field set up by some potential and may move. They also fall back into their valence band, as well. But heat continues to percolate them into the conduction band. In equilibrium, at some temperature and at any given moment in time, there will be some fairly predictable number of electrons in the conduction band. (A tightly skirted Gaussian distribution around that number, anyway.) There's more to conductivity, such as the mobility of these conduction band electrons. But that is the basic idea.

So far, I've not mentioned holes at all, nor P nor N materials.

Now think of silicon, itself. It's also a solid and exists at some thermal temperature. It's not very conductive, but it is more conductive than an insulator. So it has some number of thermally agitated electrons in the conduction band. Not so many, and perhaps their mobility is more limited than in copper for example, but they are there all the same.

For copper (and there are various forms of copper -- annealed, ETP, etc) at room temperature, the number of thermally agitated conduction band electrons per unit volume is about \$8.5\cdot 10^{22}\:\frac{\textrm{electrons}}{\textrm{cm}^3}\$. In silicon, this is about \$1.5\cdot 10^{10}\:\frac{\textrm{electrons}}{\textrm{cm}^3}\$. As you can see, a lot fewer in silicon, than in copper. Silicon is a semiconductor. And glass will have far, far fewer still (despite having silicon as part of its structure.)

In pure (intrinsic) silicon, each conduction band electron is mirrored by a hole. Think of it as a shadow. The density of conduction band electrons will be mirrored by exactly the same number of holes. Thermal agitation continues, as always, to produce more hole-electron pairs, but this is happening while other hole-electron pairs also recombine. The thermal equilibrium state will have some momentary number of hole-electron pairs that stays fairly constant.

When you add dopants to the structure, it distorts the work functions of the valence electrons in the donor atoms, which are struggling to conform to the overall bonding structure that the intrinsic and dominant silicon atoms prefer. With N dopants, the donor atoms arrange themselves to hook up to four other silicon atoms (like all the other silicon is doing), but this leaves them with a "spare" electron that is effectively "pushed" by these circumstances to be very close towards the conduction band and therefore requires quite a lot less thermal agitation to drive it into the conduction band. (Perhaps some 20 times closer, in fact, or to within 5% of it -- very close.) So these donors, because the thermal agitation energy has a lot less work to do, leading to much higher probabilities of finding one of their electrons in the conduction band, create more available electrons in equilibrium and a higher density of them. How many more depends on the concentration of the N dopant. But this doesn't really increase the number of holes, because these dopant holes are more like tiny divots than holes. This increased number of electrons in the conduction band means there is a newly increased chance that some of this larger number can fall into those much deeper and more important silicon holes (recombination), reducing their equilibrium density. So an N material will have more conduction band electrons and fewer holes. A similar, reverse argument then applies to P materials.

There is a nice mass-action law that applies here:

$$n\cdot p = n_i^2$$

Which is to say that if you know the density of conduction band electrons in an intrinsic, pure silicon block (\$n_i\$) and you can work out the new density of conduction band electrons you made by adding an N dopant to such a material (\$n\$), then you can compute the density of holes you have remaining using the above equation (\$p\$.) That's a "nice to have."

By now, you should be able to see that N type materials still have holes and that P type materials still have electrons. It's just that there's a different ratio that is no longer 1:1, is all.