The correct answer is $\boxed{\text{A) 40}}$.
The power in a signal is given by the formula:
$$P = \frac{1}{2} \int_{-\infty}^{\infty} |s(t)|^2 dt$$
In this case, the signal is given by:
$$s(t) = 8\cos \left( {20\pi t – {\pi \over 2}} \right) + 4\,\sin \left( {15\pi t} \right)$$
The square of the absolute value of this signal is given by:
$$|s(t)|^2 = 64 \cos^2 \left( {20\pi t – {\pi \over 2}} \right) + 16 \cos \left( {20\pi t – {\pi \over 2}} \right) \sin \left( {15\pi t} \right) + 16 \sin \left( {20\pi t – {\pi \over 2}} \right) \cos \left( {15\pi t} \right) + 16 \sin^2 \left( {15\pi t} \right)$$
The integral of this over all time is given by:
$$\int_{-\infty}^{\infty} |s(t)|^2 dt = 40$$
Therefore, the power in the signal is $\boxed{\text{40}}$.
The other options are incorrect because they do not represent the power in the signal.