= Solution
Choose $A^\mu=(\Phi,\mathbf A)$, so that $F_{i0}=E_i$ and $F_{ij}=\epsilon_{ijk}B_k$, with $\epsilon_{123}=1$. The definitions of the <polarization-magnetization tensor> imply $M^{0i}=-P_i$ and $M^{ij}=\epsilon_{ijk}M_k$. Separating temporal and spatial indices,
$$
\frac12M^{\mu\nu}F_{\mu\nu}=M^{i0}F_{i0}+\frac12M^{ij}F_{ij}
=\mathbf P\cdot\mathbf E+\frac12\epsilon_{ijk}\epsilon_{ij\ell}M_kB_\ell.
$$
The <Levi-Civita symbol> contraction equals $2\delta_{k\ell}$, giving
$$
\boxed{\mathcal L_{\rm int}=\mathbf P\cdot\mathbf E+\mathbf M\cdot\mathbf B.}
$$
Thus $\mathbf P$ is the <electric dipole moment> per volume, or <electric polarization>, and $\mathbf M$ is the <magnetic dipole moment> per volume, or <magnetization>. The signs are those of dipole interactions in a <Lagrangian>; the associated fixed-dipole potential energies have the opposite signs. Consistently, the equation in part (a) has components
$$
\nabla\cdot(\mathbf E+\mathbf P)=0,\qquad
\nabla\times(\mathbf B-\mathbf M)=\partial_t(\mathbf E+\mathbf P),
$$
so $\rho_b=-\nabla\cdot\mathbf P$ and $\mathbf J_b=\partial_t\mathbf P+\nabla\times\mathbf M$ are the <bound charge and bound current>.
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