Let $X \sim $ Exp$(\lambda)$ and $Y \sim$ Poisson$(\gamma)$ independence is assumed. Find $P(X<Y)$.
Would this be as follows:
$\displaystyle P(X < Y) = P(X < Y \mid Y = y)P(Y=y) = \frac{P(X < y, Y=y)}{P(Y=y)}P(Y=y) = P(X<y)P(Y=y) \\ = \displaystyle \int_0^y \lambda e^{-\lambda x }dx \frac{e^{-\gamma}\gamma^y}{y!} \\ = \displaystyle(1-e^{-\lambda y})\frac{e^{-\gamma}\gamma^y}{y!}$
If this is incorrect an explanation is much appreciated.
Your answer cannot depend on $y$.
If you condition on $Y$ as you have done here, then your probability must be computed over all outcomes of $Y$. So in particular, $$\Pr[X < Y] = \sum_{y=0}^\infty \Pr[X < Y \mid Y = y]\Pr[Y = y]$$ by the law of total probability. Then $$\Pr[X < Y \mid Y = y] = \Pr[X < y] = 1 - e^{-\lambda y}$$ because $F_X(x) = \Pr[X \le x] = 1 - e^{-\lambda x}$. It follows that $$\Pr[X < Y] = \sum_{y=0}^\infty (1 - e^{-\lambda y}) e^{-\gamma} \frac{\gamma^y}{y!}.$$ I leave it as an exercise for you to compute this sum.
Alternatively, if you condition on $X$, then $$\Pr[X < Y] = \int_{x=0}^\infty \Pr[X < Y \mid X = x]f_X(x) \, dx.$$ But this is a much less pleasant calculation since $$\Pr[X < Y \mid X = x] = \Pr[Y > x] = \sum_{y = \lfloor x \rfloor + 1}^\infty e^{-\gamma} \frac{\gamma^y}{y!}$$ and this is rather inconvenient to continue.