That is a statement about the energy, as seen by a particular observer.
Remember that the energy is an observer dependent quantity. In special relativity we defined the energy of a particle with 4-momentum $p^{\mu}$ measured by an observer with 4-velocity $u^{\mu}$ as:
$$E^{(u)} = - \eta_{\mu \nu} u^{\mu} p^{\nu} > 0$$
that in general relativity generalizes to
$$E^{(u)} = - g_{\mu \nu} u^{\mu} p^{\nu} > 0$$
For instance for a static observer in special relativity, that is $u^{\mu} = (1,0,0,0)$:
$$E^{(static)} = - p_{0}$$
For the particle to be moving forward in time, the energy must be positive. Notice that this is a tensorial statement, so it's true in every coordinate frame.
Now in the kerr spacetime
$$E^{(static)} = E$$
where $E$ is the constant of motion $-(\partial_t)^{\mu} u_{\mu} = -u_0 = -p_0$ (the last equality can always be satisfied, using the reparametrization freedom of the geodesic) associated to the timelike Killing vector $\partial_t = (1,0,0,0)$, therefore $E$ can be interpreted as the energy seen by a static observer at infinity, and must be positive.
If we are inside the ergoregion, there are not static observers, since the black holes is dragging us. A convenient observer that is corotating with the hole has four velocity $u^{\mu} \propto (1,0,0,\Omega_H)$, therefore:
$$E^{(rotating)} \propto (E-\Omega_H L)$$
where again $L$ is the constant of motion associated to the rotational Killing vector $\partial_\phi = (0,0,0,1)$. The statement that the energy seen by such an observer is positive implies the statement $p^{\mu}\chi_{\mu} < 0$.
The Kerr spacetime is peculiar since following a process of particle decay $E^{(0)} = E^{(1)} + E^{(2)}$ certain particles can have $E^{(2)} < 0$, but there is no contradiction with what I said before, since this happen only if these particles are unable to escape to infinity, therefore there isn't an interpretation as energy seen by a static observer at infinity.
Notice that all the above reasonings are done before crossing $r_+$.
