A man whose father is colorblind marries a woman who is the daughter of a colorblind mother. What will be the phenotype of their offspring?

A man is normal and his father is colorblind, but here the X chromosome is not transmitted from his father. The woman is a carrier because her mother is colorblind. So, a normal man with a carrier woman. Hence, your cross is as follows:

(X Y) x (Xc X)

Based on this crossing, from the above-given figure, we can conclude that there is a 50% chance of having a colorblind son and a 50% chance of having a carrier daughter. Therefore, the correct answer is that there will be a mix of colorblind sons and carrier daughters. The probability of each phenotype is 50%.