= Solution
There are errors in the original PDF at this point, not merely in the TeX transcription. The second exponential must denote the <eigenvalue> $\lambda_-(z)=e^{-\gamma^-z}$, not the propagation coefficient itself. More substantially, the printed prefactor $\operatorname{sech}(\delta z)$ is inconsistent with the stated exponential <eigenvalues>. It must be $\cosh(\delta z)$. The original's primed Jones-matrix label distinguishes the distributed component from the earlier static matrix; below write $\mathcal J(z)$ to avoid confusing that prime with a derivative.
Complex <eigenvalues> have no intrinsic largest/smallest ordering, so $+$ and $-$ are modal labels. Keep the circular <eigenvectors> fixed, put $\gamma=(\gamma^++\gamma^-)/2$ and $\delta=(\gamma^+-\gamma^-)/2$, and use the <spectral decomposition> from part (c):
$$
\mathcal J(z)=e^{-\gamma^+z}P_++e^{-\gamma^-z}P_-.
$$
Its diagonal entries are $(e^{-\gamma^+z}+e^{-\gamma^-z})/2=e^{-\gamma z}\cosh(\delta z)$, and its upper off-diagonal entry is $i(e^{-\gamma^+z}-e^{-\gamma^-z})/2=-ie^{-\gamma z}\sinh(\delta z)$. Therefore
$$
\boxed{\mathcal J(z)=e^{-\gamma z}
\begin{pmatrix}\cosh(\delta z)&-i\sinh(\delta z)\\i\sinh(\delta z)&\cosh(\delta z)\end{pmatrix}.}
$$
Where $\cosh(\delta z)\neq0$, this can be factored as
$$
\mathcal J(z)=e^{-\gamma z}\cosh(\delta z)
\begin{pmatrix}1&-i\tanh(\delta z)\\i\tanh(\delta z)&1\end{pmatrix}.
$$
The unfactored expression is valid even at zeros of the cosine-like factor when $\delta$ is imaginary. The printed expression is $\operatorname{sech}^2(\delta z)$ times this correct matrix; its <eigenvalues> are consequently $e^{-\gamma^\pm z}\operatorname{sech}^2(\delta z)$ rather than those specified. A second check is $\det\mathcal J=e^{-2\gamma z}$, whereas the printed version has <determinant> $e^{-2\gamma z}\operatorname{sech}^4(\delta z)$. Thus the requested printed identity cannot be proved except in the degenerate case $\delta=0$; the following parts use the consistent propagator.
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