Solution:

Since the eight cards cover all seven numbers and all seven colors, exactly one number and exactly one color appear twice. Sharon can discard a card exactly when a single card carries both the repeated number and the repeated color: that card is then the unique discardable one, while if no card carries both, removing any card loses a number or a color.

Hands of the first type consist of a rainbow set of seven cards — one of each number and each color, which is one of $7!$ permutation patterns — plus any of the remaining $42$ cards, and every such hand arises exactly once this way: $7! \cdot 42 = 211680$ hands. For the second type, choose the repeated number ($7$ ways) and the two colors of its cards ($\binom{7}{2} = 21$ ways); the repeated color must be one of the other $5$ colors, and the numbers of its two cards come from the remaining $6$ numbers ($\binom{6}{2} = 15$ ways); finally match the last four numbers to the last four colors ($4! = 24$ ways). That is $7 \cdot 21 \cdot 5 \cdot 15 \cdot 24 = 264600$ hands.

The probability is $\frac{211680}{211680 + 264600} = \frac{4}{9},$ so $p + q = 4 + 9 = 13.$