Answer
$\displaystyle \color{blue}{p_W(k) = {4\choose (k-1)/2}\left(\frac{2}{3}\right)^{(k-1)/2}\left(\frac{1}{3}\right)^{4-(k-1)/2}, \ k = 1,3,5,7,9}.$
Work Step by Step
The probability distribution function for $X$ is the binomial distribution with parameters $n=4$ and probability of success $p=2/3$ and has the form (See Theorem 3.2.1):
$\begin{align*}
p_X(x) &= P(X=x),\ x=0,1,2,3,4 \\
p_X(x) &= {4\choose x}\left(\frac{2}{3}\right)^x\left(\frac{1}{3}\right)^{4-x}, \ x = 0,1,2,3,4. \qquad \text{(Eq. 1)}
\end{align*}$
Let $W=2X+1$. Them the pdf of $W$ may be obtained as
$\begin{align*}
p_W(k) &= P(W=k) \\
&= P(2X+1=k) \\
p_W(k) &= P(X=(k-1)/2). \qquad \text{(Eq. 2)}
\end{align*}$
Now, $X$ has non-zero probabilities only for $X=0,1,2,3,4$.
Thus, only values for $k$ such that $(k-1)/2 = 0, 1, 2, 3$, or $4$ have non-zero probabilities.
Equivalently, if $k=1, 3, 5, 7$, or $9$, then $P(W=k)$ is non-zero.
Thus, continuing from Eq. 2, we have
$\begin{align*}
p_W(k) &= P(X=(k-1)/2),\ k=1, 3, 5, 7, 9 \\
\color{blue}{p_W(k)} &\color{blue}{= {4\choose (k-1)/2}\left(\frac{2}{3}\right)^{(k-1)/2}\left(\frac{1}{3}\right)^{4-(k-1)/2}, \ k = 1,3,5,7,9},
\end{align*}$
where we have susbstituted $(k-1)/2$ for $x$ in the pdf of $X$ in Eq. 1.
