Does a thrown ball have kinetic energy at the top of the curve?

The answer is wrong. Some author confused the situation when the ball is moving only vertically (and a graph as a function of time) with this case where there is horizontal motion. The horizontal component of the velocity is constant in a ballistic trajectory, it is the same at points A, B, and C.

The kinetic energy is zero only when the ball is stationary, and the ball is stationary only at E: so this is the only point where the kinetic energy is zero.

So... do not trust this book.

The book text is wrong.

If it has kinetic energy at D then it has kinetic energy at B.
There is component of motion in the X direction.

At D there is the potential energy of compressed air (what makes it bounce).

The correct answer E is not even a choice in the book.

Over much deliberation, I agree with Pieter's answer. Perhaps I shall elaborate more. The equation of kinetic energy is $\frac{1}{2}mv^2$. The motion of the ball is a projectile motion, and can be resolved by 2 vectors: the horizontal and the vertical.

The vertical component of this velocity decreases by $-g$ as it approaches the maximum height. Hence, it's velocity at the maximum height is zero, as the energy over here has become potential energy.

The horizontal component, on the other hand, is not zero, and keeps on decreasing, as energy is lost via heat energy.

Hence, our resultant velocity is not zero. Hence $\frac{1}{2}mv^2$ is definitely not zero. Hence, only E, when the ball is at rest, is correct.