The way to approach the problem initially is to consider what you know about the reaction
$$ n \longrightarrow \mathrm{p}^+ + \mathrm{e}^- + \bar{\nu}_e \,.$$
Because of the relativity principle we can consider the reaction in the rest-frame of the neutron without loss of generality and we know that both (three-)momentum and energy are conserved.
$$\begin{align*}
0 &= p_e + p_\nu + p_p \\
m_n &= E_e + E_\nu + E_p \,
\end{align*}$$
which is two equations with six unknowns, but we also have three relationships between the energy and momentum of the particles so, the system is only under-constrained by one degree of freedom. One assumption will solve the problem for us, it just has to be the right assumption for getting the electron the maximum energy.
It should be obvious that maximum energy is linked to maximum momentum, so we expect the proton and neutrino momenta to be co-linear. So let's give one all the recoil momentum and leave the other at rest. How does it shake out?
Leaving the nucleus at rest the two light particle recoil from each other, and the neutrino can be treated as ultra-relativistic, meaning the kinetic energy is
$$ T_\nu = E_\nu = \left|p_e\right| \,. \tag{1} $$
If, instead we leave the neutrino at rest, the very heavy proton recoils at non-relativistic speed (check this later) so that
$$ T_p = \frac{p_e^2}{2m_p} \,. \tag{2}$$
In these "one at rest" scenarios the more energy the other moving particle has the less the electron gets, so we need to figure which one of (1) and (2) is smallest and go with that scenario.
Two ways to proceed. The safe way is to follow each scenario all the way through, the other is to look at the scale of the quantities and try to guess. Let's start by guessing. Both have the same dimensionality and both have (at least) one factor of $p_e$. So if we know the size of the coefficients of $p_e$ we know the answer.
$$\begin{align*}
T_\nu &= (1) p_e \tag{1b} \\
T_p &= \left( \frac{p_e}{2 m_p} \right) p_e \tag{2b} \,,
\end{align*}$$
but the available energy is of order $m_n - m_p$ which is much less that $m_p$, so $$ \frac{p_e}{2m_p} \ll 1 \,,$$
and our choice is obvious. Scenario (2) (with the neutrino at rest) results in the most energy for the electron.
Going back to our system of equations we get
$$\begin{align*}
0 &= p_e - 0 - p_p \\
m_n &= E_e + m_\nu + E_p \,
\end{align*}$$
and applying the non-relativisitc approximation to the proton
$$\begin{align*}
p_p &= p_e \\
m_n &= E_e + m_\nu + m_p + \frac{p_p^2}{2m_p} p_p^2 \\
m_n - m_p &\approx E_e + 0 + \frac{p_e^2}{2m_p} \\
&= E_e + \frac{E_e^2 - m_e^2}{2m_p} \,,
\end{align*}$$
and the fraction on the RHS is small leading to $E_e \approx m_n - m_p$. So the electron ends up with essentially all the energy. This situation is even more extreme in the beta decay of a composite nucleus because the mass ratio is even larger.
Now check the "non-relativistic" assumption for the proton. The kinetic energy available is approximately
$$\begin{align*}
T_p &= \frac{p_e^2}{2m_p} \\
&= \frac{\left(m_n - m_p\right)^2 - m_e^2}{2m_p} \\
&\approx \frac{\left(939.6 - 938.2 \right)^2 - (.5)^2}{2 (938.2)} \tag{masses in MeV} \\
&= 9.11 \times 10^{-4} \,\mathrm{MeV} \,.
\end{align*}$$
Kinetic energy on order of one part in $10^7$ of mass is safely non-relativistic.
Finally, consider that the given the tiny mass of the neutrino the "neutrino at rest" scenario is essentially what was expected of beta decay as a two-body problem, Meaning that the energy we found for the electron is approximately the end-point energy of the spectrum (yeah, we did it right). You should also notice that with the neutrino momentum vanishingly small the phase space for this outcome is likewise vanishing which is part of what makes decay end-point measurements hard.
Best Answer
The solution you quote actually doesn't conserve momentum. You can use that $p^2$ is a Lorentz invariant and solve $(p_\nu+p_p)^2=(p_e+p_n)^2$, considering the left hand side in the lab frame and the right hand side in the CM frame. You can find $E_\nu$ and check that now momentum is conserved. Anyway, as mentioned in the previous comment, $E_\nu=m_e+m_n-m_p$ is a very good approximation.