Step 1: The data.
The mother has blood group B. The children are 25% A, 25% AB, and 50% B. We must find the parents' genotypes.
Step 2: Mother's genotype.
Since some children are blood group A, the mother (group B) must be able to pass an $I^O$ allele. So the mother is $I^BI^O$.
Step 3: What the father must give.
Children appear in groups A, AB and B. For an A child the father must supply $I^A$, and for the AB children he must also supply $I^A$. The presence of group B children too means $I^A$ and $I^B$ both come from the parents. Testing father $I^AI^B$ fits.
Step 4: Do the cross.
Father $I^AI^B$ crossed with mother $I^BI^O$ gives four equally likely children: $I^AI^B$ (AB), $I^AI^O$ (A), $I^BI^B$ (B), $I^BI^O$ (B).
Step 5: Count the phenotypes.
That gives 25% AB, 25% A, and 50% B (two of the four are B). This exactly matches the children described.
Step 6: Conclusion.
Therefore the correct answer is "$I^AI^B$ father and $I^BI^O$ mother". \[ \boxed{I^AI^B \text{ father, } I^BI^O \text{ mother}} \]