Measures for the degeneracy from protons? Come to think of it, the only things we hear about degeneracy is neutron and electron.
I made a generalization:


White Dwarf: Supported by electron degeneracy.
Neutron Star: Supported by neutron degeneracy and nuclear forces, and sometimes centrifugal force.
Quark Star: Most likely supported by quark degeneracy.


Now, we can see that there is no mention of proton degeneracy. 

A quick search lead to the Wikipedia subsection Degenerate matter; Proton degeneracy, and the answer that yes, there is proton degeneracy, but is much less effective than other forms of degenerate matter.

My question: What is the force exerted by degenerate protons, and what equation is it modeled by?
Stars are electrically neutral. In the optimal case for proton degeneracy, a star is made of hydrogen, maximising the proton number density. The number density of electrons is exactly the same.

Since the Fermi momentum $p_F$ of a species is dependent only on number density$^1$, the Fermi momentum of the electrons and protons are the same. Pressure however$^1$, is an increasing function of $p_F/mc$ and since the mass of an electron is 1800 times smaller, the electron degeneracy pressure is always much higher than the proton degeneracy pressure.

Degenerate protons do exist. They form of order 1% of the density in the neutron fluid region (the bulk) of a neutron star. However, they contribute much less than a thousandth of the pressure and always lower than the degenerate electron pressure.

Actually, the form of the pressure equation (see below) does actually admit the possibility that the degeneracy pressures of electrons and protons become more similar as $p_F/mc \gg 1$ (aka ultra-relativistic). But by that time, the energies of the electrons and protons are easily sufficient to create neutrons and an equilibrium is set up so that neutrons always outnumber the protons by at at least 8:1 and therefore their degeneracy pressure will dominate.

(1) For those interested: The Fermi momentum is
$$p_{F} = \left(\frac{3}{8\pi}\right)^{1/3} h n^{1/3}\ ,$$
where $n$ is the number density of fermions.

Whilst ideal fermion degeneracy pressure is given by
$$P = \frac{\pi m^4 c^5}{3 h^3} \left[x (2x^2 -3)(1 + x^2)^{1/2} + 3
  \sinh^{-1}(x) \right]\ , $$
where $x = p_F/mc$.