Answer
$\color{blue}{ \frac{32}{27}e^{-3} \;\approx\; 0.059}$
Work Step by Step
Note that
$\begin{align*}
M_W(t) &= \underbrace{e^{-3+3e^t}}_{M_X(t)}\cdot \underbrace{\left(\dfrac{2}{3}+\dfrac{1}{3}e^t\right)^4}_{M_Y(t)} \\
&= M_X(t)\cdot M_Y(t) & [ X\sim\text{Poisson}(\lambda=3), Y\sim\text{Binominal}(n=4,p={\scriptsize \frac{1}{3}})\ ]\\
&= M_{X+Y}(t), &\text{[ if}\ X,Y\ \text{are independent, Theorem 3.12.3.b ]}
\end{align*}$
So that $M_W(t)$ is the mfg of the random variable $X+Y$, where $X$ and $Y$ are independent random variables and $X\sim \text{Poisson}(\lambda=3)$ and $Y\sim \text{Binomial}(n=4,p=\frac{1}{3})$. Thus, from Theorem 3.12.2, $W=X+Y$.
Now, if $W=X+Y\le 1$, then $W=0,1,2,3,\ldots$ as $X,Y=0,1,2,3,\ldots.$
Thus, if $W\le 1$, then either
$\quad {\rm(i)}\ W=0$, or
$\quad {\rm(ii)}\ W=1$.
If (i) $W=X+Y=0$, then $X=Y=0$.
If (ii) $W=X+Y=1$, then either $X=0,Y=1$ or $X=1,Y=0$.
Thus,
$\begin{align*}
P(W\le 1) &= P(W=0) + P(W=1) \\
&= P(X=0,Y=0) + P(X=0,Y=1) + P(X=1,Y=0) \\
&= P(X=0)\cdot P(Y=0) + P(X=0)\cdot P(Y=1) + P(X=1)\cdot P(Y=0) \\
&=f_X(0)f_Y(0) + f_X(0)f_Y(1) + f_X(1)f_Y(0) \\
& \qquad\qquad {\scriptsize f_X(x) = \frac{e^{-3}3^x}{x!},\quad f_Y(y)={4\choose y}\left(\frac{1}{3}\right)^y\left(\frac{2}{3}\right)^{4-y},\ x,y=0,1,2,3,\ldots} \\
&=\frac{e^{-3}3^0}{0!}\cdot {4\choose 0}\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^4 + \frac{e^{-3}3^0}{0!}\cdot {4\choose 1}\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^3 + \frac{e^{-3}3^1}{1!}\cdot {4\choose 0}\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^4 \\
&= \frac{e^{-3}2^4}{3^4} + \frac{e^{-3}(4)(2^3)}{3^4} + \frac{e^{-3}(3)(2^4)}{3^4} \\
&= \frac{96}{3^4}e^{-3} \\
\color{blue}{P(W\le 1)}\ &\color{blue}{= \frac{32}{27}e^{-3} \;\approx\; 0.059}
\end{align*}$
