A phenotypically normal woman, whose father suffered from...

Haemophilia is an X-linked recessive disorder. Let the normal allele be $X^{H}$ and the haemophilia allele be $X^{h}$.

The genotype of the woman's father is $X^{h}Y$.

Since the woman is phenotypically normal but her father was haemophilic, she must have inherited the $X^{h}$ chromosome from him. Thus, she is a carrier and her genotype is $X^{H}X^{h}$.

The woman marries a normal man, whose genotype is $X^{H}Y$.

The cross is $X^{H}X^{h} \times X^{H}Y$.

The possible genotypes of the progeny are:
$X^{H}X^{H}$ (Normal female) - $25\%$
$X^{H}X^{h}$ (Carrier female, phenotypically normal) - $25\%$
$X^{H}Y$ (Normal male) - $25\%$
$X^{h}Y$ (Haemophilic male) - $25\%$

The only affected progeny is the haemophilic male ($X^{h}Y$).

Therefore, the percentage of the total progeny expected to be affected by haemophilia is $25\%$.

Answer: $25\%$