Ans: B
Join $CD$. Since $AC$ is a diameter, then $\angle ADC = 90^\circ$.
Since $BCDE$ is a cyclic quadrilateral, we have
$\begin{array}{rcl}
\angle CBE & = & 180^\circ – \angle ADC – \angle EDA \\
& = & 180^\circ – 90^\circ – 28^\circ \\
& = & 62^\circ
\end{array}$
