= Solution
In the rest frame $u^\mu=(1,0,0,0)$ and $h_{\mu\nu}=\operatorname{diag}(0,1,1,1)$. The two material contractions in the printed action become $-\mathbf P^2$ and $-2\mathbf M^2$. Consequently its material <Lagrangian density> is
$$
\mathcal L_{\rm mat}=\mathbf P\cdot\mathbf E+\mathbf M\cdot\mathbf B+\frac{\mathbf P^2}{2\kappa}+\frac{\mathbf M^2}{2\chi}.
$$
The <Euler-Lagrange field equations> are algebraic, since no derivatives of the material variables occur. They give
$$
\boxed{\mathbf P=-\kappa\mathbf E,\qquad \mathbf M=-\chi\mathbf B\quad\text{for the printed action}.}
$$
These also follow from the $i0$ and $ij$ components of the corrected equation in part (b). If instead the intended starting point was the PDF's displayed equation in part (b), its predictions would be
$$
\boxed{\mathbf P=-\kappa\mathbf E,\qquad \mathbf M=+\chi\mathbf B\quad\text{for the printed target equation}.}
$$
The two magnetic answers express the documented sign error, rather than different choices of frame. Here the inverse coefficients in the action require nonzero $\kappa,\chi$; zero response is obtained as a limiting constitutive law.
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