Before $\beta$ decay neutron is at rest
Hence $E_n=m_n c^2, p_n=0$ as the velocity is zero.
By the law of conservation of momentum.
$$
p_n=p_p+p_e
$$
Let $\mathrm{p}_{\mathrm{e}}=\mathrm{p}_{\mathrm{p}}$ then
$$
\mathrm{p}_{\mathrm{p}}+\mathrm{p}_{\mathrm{e}}=0 \Rightarrow\left|\mathrm{p}_{\mathrm{p}}\right|=\left|\mathrm{p}_{\mathrm{e}}\right|=\mathrm{p}
$$
Also, energy of proton
$$
\mathrm{E}_{\mathrm{p}}=\left(\mathrm{m}_{\mathrm{p}}^2 \mathrm{c}^4+\mathrm{p}_{\mathrm{p}}^2 \mathrm{c}^2\right)^{\frac{1}{2}}=\sqrt{\mathrm{m}_{\mathrm{p}}^2 \mathrm{c}^4+\mathrm{p}_{\mathrm{p}}^2 \mathrm{c}^2}
$$
energy of electron
$$
\mathrm{E}_{\mathrm{e}}=\left(\mathrm{m}_{\mathrm{e}}^2 \mathrm{c}^4+\mathrm{p}_{\mathrm{e}}^2 \mathrm{c}^2\right)^{\frac{1}{2}}=\sqrt{\mathrm{m}_{\mathrm{e}}^2 \mathrm{c}^4+\mathrm{pe}^2 \mathrm{c}^2}
$$
By the law of conservation of energy
$$
\begin{aligned}
&\begin{array}{l}
\mathrm{E}_{\mathrm{p}}+\mathrm{E}_{\mathrm{e}}=\mathrm{E}_{\mathrm{n}} \\
\text { if }\left|\mathrm{p}_{\mathrm{e}}\right|=\left|\mathrm{p}_{\mathrm{p}}\right|=\mathrm{p}
\end{array} \\
&\left(\mathrm{m}_{\mathrm{p}}^2 \mathrm{c}^4+\mathrm{p}^2 \mathrm{c}^2\right)^{\frac{1}{2}}+\left(\mathrm{m}_{\mathrm{e}}^2 \mathrm{c}^4+\mathrm{p}^2 \mathrm{c}^2\right)^{\frac{1}{2}}=\mathrm{m}_{\mathrm{n}} \mathrm{c}^2 \\
&\mathrm{~m}_{\mathrm{p}} \mathrm{c}^2 \approx 936 \mathrm{MeV}, \mathrm{m}_{\mathrm{n}} \mathrm{c}^2 \approx 938 \mathrm{MeV}, \\
&\mathrm{m}_{\mathrm{e}} \mathrm{c}^2=0.51 \mathrm{MeV} \simeq 0.5 \mathrm{MeV}
\end{aligned}
$$
As, the energy difference between $\mathrm{n}$ and $\mathrm{p}$ is very small, pc will be small, $\mathrm{pc}< < \mathrm{m}_{\mathrm{p}} \mathrm{c}^2$, while pc may be greater than $\mathrm{m}_{\mathrm{e}} \mathrm{c}^2$, So by neglecting
$$
\begin{aligned}
&\left(\mathrm{m}_{\mathrm{e}} \mathrm{c}^2\right)=(0.5)^2 \\
\Rightarrow & \mathrm{m}_{\mathrm{p}} \mathrm{c}^2+\frac{\mathrm{p}^2 \mathrm{c}^2}{2 \mathrm{~m}_{\mathrm{p}}^2 \mathrm{c}^4}=\mathrm{m}_{\mathrm{n}} \mathrm{c}^2-\mathrm{pc}
\end{aligned}
$$
or $m_p c^2+\frac{p^2 c^2}{2 m_p^2 c^4}+p c=m_n c^2$
To first order
$$
\begin{aligned}
\mathrm{pc} &=\mathrm{m}_{\mathrm{n}} \mathrm{c}^2-\mathrm{m}_{\mathrm{p}} \mathrm{c}^2 \\
&=938 \mathrm{MeV}-936 \mathrm{MeV}=2 \mathrm{MeV} \\
\mathrm{E} &=\mathrm{mc}^2, \mathrm{E}^2=\mathrm{m}^2 \mathrm{c}^4
\end{aligned}
$$
$\mathrm{E}$ is the energy of either proton or neutron, then

$$
\begin{gathered}
\mathrm{E}_{\mathrm{p}}=\left(\mathrm{m}_{\mathrm{p}}^2 \mathrm{c}^4+\mathrm{p}^2 \mathrm{c}^2\right)^{\frac{1}{2}}=\sqrt{936^2+2^2}=936 \mathrm{MeV} \\
\mathrm{E}_{\mathrm{e}}=\left(\mathrm{m}_{\mathrm{e}}^2 \mathrm{c}^4+\mathrm{p}^2 \mathrm{c}^2\right)^{\frac{1}{2}}=\sqrt{(0.51)^2+2^2}=2.06 \mathrm{MeV}
\end{gathered}
$$