Consider a Carnot heat engine and a refrigerator operating between the same two thermal energy reservoirs. If?

There is not enough information to answer this question.  Nowhere does it specify that the refrigerator is a perfect Carnot refrigerator.

Gintable is being a bit pedantic with you. He sometimes does that in an attempt to score a "Best Answer" with no mental effort and still somehow look "clever". Perhaps you've no idea how important and prestigious the Best Answer Mafia is. As with any corruption, it's the poor (questioners) who suffer from this behaviour and of course it lowers the general standard of many responses to near rock bottom.

Anyhow let's assume what's actually fairly clear, that both of your machines are using Carnot cycles and operating between the same two temperatures. The heat engine has thermal efficiency (1-Tc/Th) = 0.8 giving Tc/Th = 0.2 So the work done by the engine from a quantity Qh of heat removed from the hot reservoir is 0.8 x Qh.

The work to be done by a refrigerator to bring Qh to the hot reservoir
= W = (1-Tc/Th).Qh = 0.8 x Qh so since W is part of Qh the heat picked up from the cold reservoir is Qc = 0.2 x Qh the COP is then:

COP = Qc/W  = 0.2 Qh/(0.8 x Qh) = 0.25