Start with the blackbody. It absorbs any light that hits it. It doesn't reflect any light, it absorbs it all. It emits as a blackbody based on it's temperature. (sorry, would have to look up the equation for that spectrum).
Now wrap this blackbody with a magic green mirror.
If the magic green mirror reflects all green light in both directions, then any green component of the blackbody radiation will be reflected back and absorbed. In order to radiate the same amount of energy as before, the temperature of the black body must rise slightly.
So outside, the spectrum would indicate a slightly hotter black body, but would have a notch in it at the green wavelengths, where the mirror is reflecting back. (and don't let anyone shine green light at it. That would confuse the issue.)
But maybe your question has the magic mirror only reflecting green light back to the observer, that would otherwise land on the black body. Let's imagine it is a supa-magic mirror that only reflects from one direction, but lets all light from the blackbody through.
IF the black-body was being heated only by a green laser (you didn't specify why it might have a spectrum, how it is getting its heat. So there are a lot of ifs and scenarios here), that would now be reflected, and the black-body temperature would go to zero, and it would not emit any wavelengths. The spectrum coming back from it would only be the green laser, reflected back by the magic mirror.
Finally: if this body is green, that is, reflects some spectrum of light, and absorbs the rest, an observer will see its black-body spectrum, but with an added component that is green if there is ambient illumination with green in it.
This is not exotic at all. Everything with a temperature emits black-body radiation. Most things also reflect back a portion of any light that lands on them. So when you measure a thing's black-body spectrum, don't shine a light on it!