Answer: D

Small Hint:

There are $20$ allowed values for each of $a$ and $b;$ $(x + iy)^4$ is real iff $x = 0,$ $y = 0,$ or $x = \pm y$

Big Hint:

Translate those into $\cos(a\pi) = 0,$ $\sin(b\pi) = 0,$ or $\cos(a\pi) = \pm\sin(b\pi)$

Solution:

There are $20$ possible values for each of $a$ and $b.$ In reduced form, denominators $1,2,3,4,5$ contribute respectively $2,2,4,4,8$ values in $[0,2),$ for a total of $20.$

Writing $x = \cos(a\pi)$ and $y = \sin(b\pi),$ the fourth power $(x + iy)^4$ is real if and only if $x = 0,$ $y = 0,$ or $x = \pm y.$

The case $x = 0$ means $a \in \left\{\dfrac12, \dfrac32\right\},$ giving $2\cdot 20 = 40$ pairs; the case $y = 0$ means $b \in \{0, 1\},$ giving another $40$ pairs, of which $4$ were already counted.

For the remaining condition $\cos(a\pi) = \pm\sin(b\pi),$ with neither side zero, the allowed values of $b$ are $\dfrac14,$ $\dfrac12,$ $\dfrac34,$ $\dfrac54,$ $\dfrac32,$ and $\dfrac74.$ The corresponding numbers of allowed values of $a$ are respectively $4,2,4,4,2,4.$ For example, when $b = \dfrac14,$ the values are $a = \dfrac14,$ $\dfrac34,$ $\dfrac54,$ and $\dfrac74;$ the other rows follow from the same quarter-turn identities. Hence this condition contributes $4+2+4+4+2+4=20$ more pairs.

In all there are $40 + 40 - 4 + 20 = 96$ valid pairs out of $400,$ so the probability is $\dfrac{96}{400} = \dfrac{6}{25}.$

Thus, the correct answer is D.