Recovering Maxwell-Boltzmann distribution from ideal-gas law

I'm reading the Blundell) Concepts in Thermal Physics_2thed Chap6 Pressure part. Before this, author shows that if we ignore the internal energy of molecules and just consider kinetic energy $$\frac{1}{2}mv^2$$ part alone with Boltzmann factor, the velocity distribution function is proportional to a Boltzmann factor and we could discover the Maxwell-Boltzmann function. It's quite interesting for me to see that using such distribution function, there's a way to derive ideal gas equation $$pV=Nk_{B}T$$.(It is dervied on Blundell textbook page.58-59)

I think my question is quite subtle, but does this imply that ideal gas behavior is totally consistent with Maxwell-Boltzmann distribution? I mean that I do not derive it in a reverse direction, from ideal-gas law to Maxwell-Boltzmann distribution. So I wonder either those things are considered to be equivalent or one may be more general than the other. Or by adding some assumptions, do those two relations become equivalent manner?

The answer to your question is quite interesting: the ideal gas equation of state is very general; it applies whenever the particles don't interact with one another (except by very short range forces to allow collisions). The relation between momentum and energy (called the dispersion relation when you do the analysis using statistical mechanics) can be anything at all and you still get $$p V = N k_B T$$.
In the Boltzmann factor it is always the energy that appears, so when the relationship between momentum and energy is different (e.g. in a relativistic gas) the distribution over momentum is different. For example one gets $$f(p) \propto p^2 e^{-E/k_B T}$$ where $$p = \gamma m v$$ and $$E = \gamma m c^2$$ and $$\gamma = 1/\sqrt{1-v^2/c^2}$$. This distribution is not the MB distribution but remarkably you still get $$p V = N k_B T$$. The easiest way to show this is using statistical mechanics via the single particle partition function.