= Solution
Use $\mathbf p=-i\hbar\nabla$ and regard $e$ as the signed charge in the printed <minimal coupling>. Multiplying the first <Dirac equation> by $c\gamma^0$ gives $H=c\boldsymbol\alpha\cdot\boldsymbol\pi+\beta mc^2+e\Phi$, with $\boldsymbol\pi=\mathbf p-e\mathbf A/c$. Removing the rest-energy <phase> yields
$$
\boxed{\begin{aligned}
i\hbar\partial_t\tilde\varphi&=e\Phi\tilde\varphi+c\boldsymbol\sigma\cdot\boldsymbol\pi\,\tilde\chi,\\
i\hbar\partial_t\tilde\chi&=c\boldsymbol\sigma\cdot\boldsymbol\pi\,\tilde\varphi+(e\Phi-2mc^2)\tilde\chi.
\end{aligned}}
$$
For the alternative equation use its specified field conventions literally. They give
$$
F_{\mu\nu}\gamma^\mu\gamma^\nu
=\frac2c\boldsymbol\alpha\cdot\mathbf E+2i\boldsymbol\Sigma\cdot\mathbf B,
\qquad \boldsymbol\Sigma=\operatorname{diag}(\boldsymbol\sigma,\boldsymbol\sigma).
$$
Consequently its <Hamiltonian> is $c\boldsymbol\alpha\cdot\mathbf p+\beta mc^2-\kappa\beta\boldsymbol\Sigma\cdot\mathbf B+(i\kappa/c)\beta\boldsymbol\alpha\cdot\mathbf E$. The two shifted equations are
$$
\boxed{\begin{aligned}
i\hbar\partial_t\tilde\varphi&=-\kappa\boldsymbol\sigma\cdot\mathbf B\,\tilde\varphi
+\left(c\boldsymbol\sigma\cdot\mathbf p+\frac{i\kappa}{c}\boldsymbol\sigma\cdot\mathbf E\right)\tilde\chi,\\
i\hbar\partial_t\tilde\chi&=\left(c\boldsymbol\sigma\cdot\mathbf p-\frac{i\kappa}{c}\boldsymbol\sigma\cdot\mathbf E\right)\tilde\varphi
+\left(-2mc^2+\kappa\boldsymbol\sigma\cdot\mathbf B\right)\tilde\chi.
\end{aligned}}
$$
These electric-field factors depend on the source's convention $E_i=cF_{0i}$; substituting a different <Gaussian> or SI definition without changing the equation would give incorrect <powers> of $c$.
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