Let us first find the boost of the momentum four-vector ($p^\mu$). In the particle rest frame, $p^\mu$ is given by, $p^\mu = (m, \textbf{0)}$. Now we go to a boosted frame in the 3-direction. In the boosted frame, we define, $p^\mu = (E, \textbf{p})$. Then we have,
%
\begin{align}
\begin{pmatrix}
E\\
p^3
\end{pmatrix}
= \mathrm{exp}\left[\eta
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\right]
\begin{pmatrix}
m\\
0
\end{pmatrix}
\end{align}
Let us define, $A =
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$.
%
Then we have,
%
\begin{align}
A^2 &=
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
= \begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
= I_{2}\\
%
\implies A^3 &= A^2 \cdot A = I_{2} A = A\\
%
\implies A^{2n} &= I_{2}\\
%
\implies A^{2n +1} &= A
\end{align}
%
\begin{align}
\mathrm{exp}\left[\eta
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\right]
&= e^{\eta A} = I_{2} + \eta A + \frac{\eta^2 A^2}{2!} + \frac{\eta^3 A^3}{3!} + \frac{\eta^4 A^4}{4!} + \frac{\eta^5 A^5}{5!} + \cdots\nonumber\\
%
&= I_{2} + \eta A + \frac{\eta^2}{2!} I_{2} + \frac{\eta^3 A}{3!} + \frac{\eta^4}{4!} I_{2} + \frac{\eta^5 A}{5!} + \cdots\nonumber\\
%
&= I_{2} \left(\frac{\eta^2}{2!} + \frac{\eta^4}{4!} + \cdots \right) + A \left(\eta + \frac{\eta^3}{3!} + \frac{\eta^5}{5!} + \cdots \right)\nonumber\\
%
&= I_{2} \cosh{\eta} + A \sinh{\eta}\nonumber\\
%
&= \cos{\eta}
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
+ \sin{\eta}
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\end{align}
Then we have,
\begin{align}
\begin{pmatrix}
E\\
p^3
\end{pmatrix}
&= \left[\cos{\eta}
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
+ \sinh{\eta}
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\right]
\begin{pmatrix}
m\\
0
\end{pmatrix}
=
\begin{pmatrix}
m \cosh{\eta}\\
m\sinh{\eta}
\end{pmatrix}
\end{align}
where, $\eta$ is the rapidity. So we have, $E = m \cosh{\eta}$ and $p^3 = m \sinh{\eta}$. Now we look at boost of the column vector, $u(p) = \sqrt{m} \begin{pmatrix}
\xi\\
\xi
\end{pmatrix}$. While considering the boost along 3-direction, we have the boosting matrix given by,
\begin{align}
\Lambda_{\frac{1}{2}} &= \mathrm{exp}\left(-\frac{i}{2} \omega_{\mu\nu} S^{\mu\nu}\right) = \mathrm{exp}\left(-\frac{i}{2} \omega_{03} S^{03}\right)\nonumber\\
%
&= \mathrm{exp}\left[-\frac{1}{4} \omega_{03}
\begin{pmatrix}
\sigma^3 & 0\\
0 & - \sigma^3
\end{pmatrix}\right]
= \mathrm{exp}\left[-\frac{\eta}{2}
\begin{pmatrix}
\sigma^3 & 0\\
0 & - \sigma^3
\end{pmatrix}\right]
\end{align}
%
where, we defined, $\eta = \omega_{03}/2$. So, the boosted version of $u(p)$ is,
%
\begin{align}
u(p) &= \Lambda_{\frac{1}{2}} \sqrt{m}
\begin{pmatrix}
\xi\\
\xi
\end{pmatrix}
= \mathrm{exp}\left[-\frac{\eta}{2}
\begin{pmatrix}
\sigma^3 & 0\\
0 & - \sigma^3
\end{pmatrix}\right] \sqrt{m}
\begin{pmatrix}
\xi\\
\xi
\end{pmatrix}
\end{align}
Now, we note that, $\sigma^3 =
\begin{pmatrix}
1 & 0\\
0 & - 1
\end{pmatrix}
\implies (\sigma^3)^2 =
\begin{pmatrix}
1 & 0\\
0 & - 1
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
0 & - 1
\end{pmatrix}
=\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix} =I_{2}
\implies (\sigma^3)^{2n} = I_{2}$ and $(\sigma^{3})^{2n + 1} = \sigma^{3}$. Proceeding as before, by defining $B = \begin{pmatrix} \sigma^3 & 0\\ 0 & - \sigma^3 \end{pmatrix}$, we can prove,
\begin{align}
\mathrm{exp}\left[-\frac{\eta}{2}
\begin{pmatrix}
\sigma^3 & 0\\
0 & - \sigma^3
\end{pmatrix}\right] = \cosh{(\eta/2)}
\begin{pmatrix}
I_{2} & 0 \\
0 & I_{2}
\end{pmatrix} - \sinh{(\eta/2)}
\begin{pmatrix}
\sigma^3 & 0 \\
0 & - \sigma^3
\end{pmatrix}
\end{align}
%
Hence we find,
%
\begin{align}
u(p) &= \mathrm{exp}\left[-\frac{\eta}{2}
\begin{pmatrix}
\sigma^3 & 0\\
0 & - \sigma^3
\end{pmatrix}\right] \sqrt{m}
\begin{pmatrix}
\xi\\
\xi
\end{pmatrix}\nonumber\\
&= \left[\cosh{(\eta/2)}
\begin{pmatrix}
I_{2} & 0 \\
0 & I_{2}
\end{pmatrix} - \sinh{(\eta/2)}
\begin{pmatrix}
\sigma^3 & 0 \\
0 & - \sigma^3
\end{pmatrix}\right] \sqrt{m}
\begin{pmatrix}
\xi\\
\xi
\end{pmatrix}\nonumber\\
&= \left[
\begin{matrix}
I_{2}\cosh{(\eta/2)} - \sigma^3 \sinh{(\eta/2)} & 0\\
0 & I_{2} \cosh{(\eta/2)} + \sigma^3 \sinh{(\eta/2)}\\
\end{matrix}\right] \sqrt{m}
\begin{pmatrix}
\xi\\
\xi
\end{pmatrix}\nonumber\\
&= \left[
\begin{matrix}
\left(\frac{e^{\eta/2} + e^{-\eta/2}}{2}\right) I_{2} - \left(\frac{e^{\eta/2} - e^{-\eta/2}}{2}\right) \sigma^3 & 0\\
0 & \left(\frac{e^{\eta/2} + e^{-\eta/2}}{2}\right) I_{2} + \left(\frac{e^{\eta/2} - e^{-\eta/2}}{2}\right) \sigma^3\\
\end{matrix}\right] \sqrt{m}
\begin{pmatrix}
\xi\\
\xi
\end{pmatrix}\nonumber\\
&= \left[
\begin{matrix}
e^{\eta/2} \left( \frac{I_{2} - \sigma^3}{2}\right) + e^{-\eta/2} \left( \frac{I_{2} + \sigma^3}{2}\right) & 0\\
0 & e^{\eta/2} \left( \frac{I_{2} + \sigma^3}{2}\right) + e^{-\eta/2} \left( \frac{I_{2} - \sigma^3}{2}\right)\\
\end{matrix}\right] \sqrt{m}
\begin{pmatrix}
\xi\\
\xi
\end{pmatrix}\nonumber\\
&= \left[
\begin{matrix}
\left(\sqrt{m} e^{\eta/2}\right) \left( \frac{I_{2} - \sigma^3}{2}\right) + \left(\sqrt{m} e^{-\eta/2}\right) \left( \frac{I_{2} + \sigma^3}{2}\right)\\
\left(\sqrt{m} e^{\eta/2}\right) \left( \frac{I_{2} + \sigma^3}{2}\right) + \left(\sqrt{m} e^{-\eta/2}\right) \left( \frac{I_{2} - \sigma^3}{2}\right)\\
\end{matrix}\right]
\begin{pmatrix}
\xi\\
\xi
\end{pmatrix}\nonumber\\
&= \left(
\begin{matrix}
\left[\sqrt{E + p^3} \left( \frac{I_{2} - \sigma^3}{2}\right) + \sqrt{E - p^3} \left( \frac{I_{2} + \sigma^3}{2}\right)\right]\xi\\
\\
\left[\sqrt{E + p^3} \left( \frac{I_{2} + \sigma^3}{2}\right) + \sqrt{E - p^3} \left( \frac{I_{2} - \sigma^3}{2}\right)\right]\xi\\
\end{matrix}\right)
\end{align}
where, we have used the following facts, $\sqrt{m}\, e^{\eta/2} = (m e^\eta)^{1/2} = \sqrt{m \cosh{\eta} + m \sinh{\eta}}$ and, $\sqrt{m}\, e^{-\eta/2} = (m e^{-\eta})^{1/2} = \sqrt{m \cosh{\eta} - m \sinh{\eta}}$.
