A man whose father is a colourblind,
marry a woman, who is a daughter of
colourblind mother. The offspring of this
couple will be
(a) all daughter and sons are colourblind
(b) 50% colourblind and 50% normal son
(c) carrier normal daughter and colourblind sons
(d) colourblind sons and normal daughter

Please explain the solution in detail.

carrier normal daughter and colourblind sons

Color blindless is an X-linked recessive disorder. Hence, the sex chromosomes (X and Y) are used when doing the punnet square.

-(c) can be used to represent color blindless, because it is the recessive trait.
-(C) can be used to represent the dominant allele for normal vision.

A man who is colorblind can be represented as (Xc Y)
A female with normal vision can be represented as (XC XC)

Hence, your cross is as follows:

(Xc Y) x (XC XC)

Based on this crossing, you can conclude that there is a 0% chance of producing a son who is colorblind. All males will have normal vision (XC Y)

From this crossing, you can also conclude that 100% of the females will be carriers (XC Xc), and thus have normal vision, because the allele for normal vision dominates the recessive allele for color blindness