Which form of the geometric distribution are you using: $\displaystyle \Pr(X = k) = (1-p)^{k-1}p$ for k = 1, 2, ..., or $\displaystyle \Pr(X=k) = (1 - p)^kp$ for k = 0, 1, ...?
In the first case, Pr(X = Y) = Pr(X = 1 /\ Y = 1) + Pr(X = 2 /\ Y = 2) + ... = Pr(X = 1)Pr(Y = 1) + Pr(X = 2)Pr(Y = 2) + ... since X and Y are independent. I believe this series is a geometric series.