The test to see whether degeneracy pressure is going to be significant is to compare $kT$ with the Fermi energy $E_F$
The Fermi energy is the energy level up to which all energy states would be occupied in a completely degenerate fermion gas. It is given by (for non-relativistic conditions)
$$ E_F = frac{h^2}{2m}left(frac{3}{8pi}right)^{2/3} n^{2/3},$$
where $m$ and $n$ are the mass and number density of the fermions (in this case, electrons).
If $E_F gg kT$, then the gas can be considered completely degenerate and the electrons exert a non-relativistic degeneracy pressure. If $E_F sim 10 kT$ the electrons are partially degenerate.
In both these circumstances, the pressure exerted by the electrons is much higher than they would have in a perfect gas, because a large fraction of electrons occupy high energy (and momentum) states.
Degeneracy can be achieved either by having a low temperature or a high number density of fermions. In a white dwarf, electrons are completely degenerate because the number density of electrons is extremely high. In a gas giant, the electron density is nowhere near as high, but then the temperature is a lot lower and so they reach a state of partial degeneracy.
The pressure of a partially degenerate gas is higher than that of a perfect gas at the same density and temperature. More importantly, the pressure is only very weakly dependent on temperature. As a gas giant radiates away its potential energy, it is able to do so whilst only contracting very slightly and the core does not become hot enough to initiate nuclear fusion.
Note that degeneracy pressure and thermal pressure are not two different things. Both arise merely as a consequence of particles having a distribution of momenta and obeying (in this case) Fermi-Dirac statistics to determine how they occupy the possible energy states. The expression $P = nkT$ is simply a convenient approximation for fermions that holds only when $E_F ll kT$ and the quantum nature of the particles is not apparent.
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