I will change notation a little to make things look more familiar. For $n=0$, $1$, $2$, $3$, and so on, we have $P(X=n)=c\frac{3^n}{n!}.$
The probabilities must add to $1$, so $\sum_{n=0}^\infty c\frac{3^n}{n!}=c\left(1+\frac{3}{1!}+\frac{3^2}{2!}+\frac{3^3}{3!}+\frac{3^4}{4!}+\cdots\right)=1.\tag{$1$}$ Recall from calculus the power series (Taylor series, Maclaurin series) for the exponential function. $e^t=\sum_{n=0}^\infty \frac{t^n}{n!}=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots.$ So (taking $t=3$) we may rewrite $(1)$ as $ce^3=1,$ from which we conclude that $c=\frac{1}{e^3}=e^{-3}$.