Start directly from Maxwell's equations in a conducting Earth:
\[\nabla\times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}, \qquad \nabla\times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}\]
For typical MT frequencies (roughly \(10^{-4}\) to \(10^{4}\) Hz) and Earth conductivities (roughly \(10^{-4}\) to \(10^{2}\) S/m), the ratio of displacement current to conduction current, \(\omega\epsilon/\sigma\), works out to about \(10^{-6}\) or smaller. This lets us drop the \(\partial\mathbf{D}/\partial t\) term entirely, turning the induction equation into a purely diffusive one rather than a wave equation. This directly falsifies option (B): displacement currents are, in fact, neglected, not retained.
Because the ionospheric current systems that generate the natural field sit hundreds of kilometres above the surface - far beyond the skin depths of interest - the incident field looks, locally, like a uniform plane wave striking the Earth from above. This validates (A).
Since the Earth is a passive, source-free medium at these frequencies (it carries no internal EMF), all the field energy is externally imposed and is dissipated resistively; the Earth never creates energy of its own. This validates (D).
Finally, combining \(\nabla\cdot\mathbf{J}=-\partial\rho_f/\partial t\) with Ohm's law \(\mathbf{J}=\sigma\mathbf{E}\) shows that any free charge inside a conductor decays with a relaxation time \(\epsilon/\sigma\) that is vanishingly small for Earth materials - charge cannot accumulate, so the medium behaves as a pure Ohmic conductor. This validates (C).
Hence the valid assumptions are (A), (C) and (D) - options 1, 3 and 4.