What is this black hole merger orbital speed? If space craft nears a black hole we see it getting slower and slower to the point it would appear to stop moving due to red shift. Then how fast are these black holes moving?

 A: Let me begin by stressing that this question does not have a physically meaningful answer in any precise way. This is due to three reasons:


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*There are no global frames of reference in general relativity (see e.g. this Q&A: How do frames of reference work in general relativity, and are they described by coordinate systems?). Hence there is no meaningful way of assigning a velocity relative to the observer (or the center-of-mass for that matter) to the black holes. In particular, if the black holes were pointlike objects (but see 2.) they would be at rest in their own local reference frame.

*The black holes are not physical objects with any precise localisation. Their individual event horizons are extended (and deformed). Hence we cannot assign a precise position to the black holes, let alone a velocity.

*The spacetime of a binary black hole merger is highly dynamical. This excludes many approximations one could make to mitigate the above two issues, strengthening their consequences.
That being said we can have a go at giving an approximate answer. Based on the relative sizes of the two photon spheres in the video, we can estimate the ratio of the two black hole masses to be roughly 1:4. Furthermore, lets assume the individual black holes are not spinning. From simulation data it can be read off (e.g. from the peak of the gravitational wave frequency) that the merger happens at angular velocity $\omega = 0.16 c^3/(GM)$, where $G$ is the gravitational constant, $c$ the speed of light, and $M$ the total mass of the binary. From Effective One Body descriptions it is known that the merger happens there and about a radial separation from the "center of mass" $r = 3 GM/c^2$. We can thus roughly estimate the orbital velocity as $r\omega =0.6\,c$.
However, I stress again that this is (and fundamentally can be) only a "back of the envelope" calculation.
