jmc

We can use complementary counting, by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$. Imagine that the $22$ other (indistinguishable) people are already seated, and fixed into place. We will place $A$, $B$, and $C$ with and without the restriction. There are $22$ places to put $A$, followed by $21$ places to put $B$, and $20$ places to put $C$ after $A$ and $B$. Hence, there are $22\cdot 21\cdot 20$ ways to place $A,B,C$ in between these people with restrictions. Without restrictions, there are $22$ places to put $A$, followed by $23$ places to put $B$, and $24$ places to put $C$ after $A$ and $B$. Hence, there are $22\cdot 23\cdot 24$ ways to place $A,B,C$ in between these people without restrictions. Thus, the desired probability is $1-\frac{22\cdot 21\cdot 20}{22\cdot 23\cdot 24}=1-\frac{420}{552}=1-\frac{35}{46}=\frac{11}{46}$, and the answer is $11+46=\boxed{57}$.