Perhaps this is too deep for me, so I don't expect you to reiterate everything again as I probably won't get it. However could you just tell me what happens in these situations?Interesting question. First, note that if this argument is valid, it will hold for any object falling into the hole, not just a photon. (Although there is a key difference between a photon and a massive object--see below.)
However, as it stands, it can't be valid, because momentum is frame-dependent, and conservation laws can't be frame-dependent. So whatever is conserved, it can't be momentum by itself; at the very least, we have to pick a particular frame in which to analyze the problem.
If you look at the actual math of freely falling objects in the gravitational field of a black hole (or indeed any spherically symmetric mass), there are two constants of the motion, energy at infinity and angular (not linear) momentum. To simplify things we can restrict ourselves to scenarios where the object is moving purely radially, so its angular momentum is zero. But even then, we have only energy at infinity as a constant of the motion (i.e., conserved quantity), and energy at infinity is evaluated in the frame in which the black hole is at rest. In that frame, the hole does not gain any momentum, by definition; it only gains energy--the energy it gains is equal to the energy at infinity of the object that falls in, so total energy in this frame is conserved.
Now what about momentum in the black hole's rest frame? If we drop in a massive object, we can drop it in "from rest at infinity", i.e., with zero initial momentum in the black hole's rest frame; in this case, total momentum is obviously conserved in the hole's rest frame, since the hole's final momentum in that frame is zero.
However, we can't drop a photon into the hole from rest at infinity, because a photon can never be at rest; this is the key difference I referred to above. What this tells me is that we can't just consider the photon in isolation; we have to consider its source as well, including the source's momentum. First take the simple case of a source that is at rest at infinity. Then when the source shoots the photon at the black hole, it recoils, so the total momentum at infinity of source + photon remains zero in the hole's rest frame. Thus, total momentum in that frame remains conserved. It is true that the source can increase the distance between itself and the hole, giving the appearance of a repulsive force exerted on the hole, by repeatedly firing photons in this way; but that is just another way of saying that we can make a photon rocket.
If the photon source has some nonzero original momentum in the black hole's rest frame, then a similar analysis to the above indicates that that will be the final total momentum of the entire system in that frame.
A conventional rocket fires its thrust into a black hole- do they both experience the same repulsive force?
A "photon rocket" fires its thrust into a black hole- do they both experience the same repulsive force?
(in both of these situations consider that the rocket and BH and a third party observer are stationary with respect to each other at t = 0, and that the forces are measured at the first instant t = 0, and that we are viewing from the frame of the third party observer who is stationary and at a finitite distance.)