Answer
$\displaystyle \color{blue}{e^{-\lambda + \lambda e^t}}$
Work Step by Step
$\begin{align*}
M_X(t) &= E(e^{Xt}) \\
&= \displaystyle \sum_{-\infty}^\infty e^{kt}P(X=k) \\
&= \displaystyle \sum_{k=0}^\infty e^{kt}p_X(k) & \text{[ since }\ p_X(k) = 0, k\not\in \{0,1,2,3,\dots\}\ ] \\
&= \displaystyle \sum_{k=0}^\infty e^{kt}\cdot\left(\frac{e^{-\lambda}\lambda^k}{k!}\right) & \text{[ since }\ p_X(k) = \frac{e^{-\lambda}\lambda^k}{k!}, k\in \{0,1,2,3,\dots\}\ ] \\
&= \displaystyle e^{-\lambda} \cdot \underbrace{\sum_{k=0}^\infty \frac{(\lambda e^t)^k}{k!}}_{e^{\lambda e^t}} & \text{[ since }\ e^r = \sum_{k=0}^\infty \frac{r^k}{k!},\ \text{let}\ r = \lambda e^t\ ]\\
&= \displaystyle e^{-\lambda} \cdot e^{\lambda e^t} \\
\color{blue}{M_X(t)}\ &\color{blue}{= e^{-\lambda + \lambda e^t}}
\end{align*}$
