We can find the number of extra clubs in the deck from the given probability, since the probability of drawing a diamond and a club is
[tex]\dfrac{\binom{13}1\binom{13+c}1}{\binom{59}2}=\dfrac{13(13+c)}{1711}[/tex]
but this would imply that [tex]c=-4[/tex], which suggests we're taking 4 clubs out of the original deck and contradicts the problem statement that there are "too many clubs".
Perhaps the question is providing the probability of drawing a diamond, THEN a club, or vice versa. In that case, we have
[tex]\dfrac{\binom{13}1}{\binom{59}1}\cdot\dfrac{\binom{13+c}1}{\binom{58}1}=\dfrac{13(13+c)}{3422}[/tex]
and solving gives [tex]c=5[/tex], which makes more sense. Then the number of extra spades in the deck must be 2.