= Solution
Choose the active <Lorentz boost>[boost] producing $p_z=m\sinh\rho$ and $E=m\cosh\rho$. The supplied two-component <Lorentz boost>[boost] $L=\exp(-\rho\sigma_3/2)$ acts on the left-chiral block $\eta$; the right-chiral block $\xi$ transforms by its inverse. Applying them to <5/b> gives
$$
\boxed{\xi(p)=\sqrt m\,e^{\rho\sigma_3/2}\chi,
\qquad \eta(p)=\sqrt m\,e^{-\rho\sigma_3/2}\chi.}
$$
Use $me^{\pm\rho}=E\pm p_z$ and $\sigma_3\chi_\pm=\pm\chi_\pm$ to obtain
$$
\boxed{u_+(p)=\begin{pmatrix}\sqrt{E+p_z}\\0\\\sqrt{E-p_z}\\0\end{pmatrix},\qquad
u_-(p)=\begin{pmatrix}0\\\sqrt{E-p_z}\\0\\\sqrt{E+p_z}\end{pmatrix}.}
$$
Both are <particle> <spinors>, with spin respectively parallel and antiparallel to the positive $z$ axis. A compact equivalent expression is $\xi=\sqrt{E+p_z\sigma_3}\chi$, $\eta=\sqrt{E-p_z\sigma_3}\chi$, using the positive matrix square roots.
The products of corresponding nonzero upper and lower components equal $\sqrt{E^2-p_z^2}=m$, so $\bar u_\pm u_\pm=2m$ and $u_\pm^\dagger u_\pm=2E$. Substitution also verifies $(E-p_z\sigma_3)\xi=m\eta$ and $(E+p_z\sigma_3)\eta=m\xi$. These checks fix which <Lorentz boost>[boost] belongs to which component and prevent silently exchanging the chiral ordering defined by the source's $U$.
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