How is $ai^{bi}$ a real number?

The number $i^{i}$ is real, which immediately implies the result you're interested in. "Proof" : the complex conjugate of a complex number is obtained by replacing every occurrence of $i$ with $-i$; so the complex conjugate of $i^{i}$ is $(-i)^{-i}=(-1/i)^{i}=i^{i}$; but the only way that complex conjugation can give back the same number is if the number is actually real.

Addendum: strictly speaking, raising imaginary numbers to imaginary powers is not well-defined. Consider the following. We have $i=e^{i\pi/2}=e^{5i\pi/2}.$ Hence $i^{i}=e^{-\pi/2}=e^{-5\pi/2},$ but these two real numbers are clearly different. This is a delicate issue, in that resolving this contradiction requires choosing a "branch" of the complex logarithm.