= Solution
Expand the original <Spinor>[two-spinor] as $\chi=c_+\chi_++c_-\chi_-$. <Orthogonality> gives
$$
c_+=\frac{\sqrt5}{3}\cos\frac\theta2+\frac23e^{-i\varphi}\sin\frac\theta2,\qquad
c_-=-\frac{\sqrt5}{3}e^{i\varphi}\sin\frac\theta2+\frac23\cos\frac\theta2.
$$
Let $r=|\mathbf p|/(E+m)$ and $w_\pm=(\chi_\pm,-\chi_\pm)^{\mathsf T}/\sqrt2$. These $w_\pm$ are <Normalization>[normalized], <Chirality>[left-chiral] <helicity> <eigenvectors>. Then
$$
\boxed{\psi_L=\sum_{\zeta=\pm1}\frac N{\sqrt2}c_\zeta(1-\zeta r)w_\zeta\,e^{i(\mathbf p\cdot\mathbf r-Et)}.}
$$
The two contributions have squared norms
$$
\boxed{\|\psi_{L,+}\|^2=|c_+|^2\frac{1-\beta}{2},\qquad
\|\psi_{L,-}\|^2=|c_-|^2\frac{1+\beta}{2},\qquad \beta=|\mathbf p|/E.}
$$
These are weights relative to the original <Normalization>[normalized] state; divide by their sum if conditioning on a <Normalization>[normalized] <Chirality>[left-chiral] projection. In the ultrarelativistic limit $N\to1/\sqrt2$, $r\to1$, the <Helicity>[positive-helicity] <Probability amplitude>[amplitude] is suppressed as $m/(2E)$, while the <Helicity>[negative-helicity] <Probability amplitude>[amplitude] tends to $c_-$. Hence
$$
\boxed{\psi_L\longrightarrow c_-w_-e^{i(\mathbf p\cdot\mathbf r-Et)}.}
$$
For <momentum> specifically along $+z$, $c_+=\sqrt5/3$ and $c_-=2/3$: the original <helicity> <probabilities> are $5/9,4/9$, the <Chirality>[left-chiral] <Norm> is $1/2-\beta/18$, and its limiting weight is $4/9$. The source does not specify that <momentum> direction, so these numerical <probabilities> are a specialization, not the general answer.
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