Answer
$\color{blue}{p_X(2x-4) = \displaystyle {4\choose x}(2/3)^x(1/3)^{4-x},\ x=0,1,2,3,4}$
Work Step by Step
If the particle in 3.3.7 is twice as likely to go right (R) than go left (L), then the random variable (r.v.) $Y$ denoting the number of R steps taken by the particle in its four-step motion will now have a binomial distribution with success (R step) probability of $2/3$ and failure (L step) probability of $1/3$ instead of $P(R)=P(L)=1/2$. The number of trials is still $n=4$ as in 3.3.7. Consequently, the pdf of $Y$ is now given by
$\displaystyle p_Y(x) = {4\choose x}(2/3)^x(1/3)^{4-x},\ x=0,1,2,3,4. \qquad \text{(Eq. 1)}$
As in 3.3.7, if the particle takes exactly $x$ R steps ($x=0,1,2,3$, or $4$) in its four-step motion then
i) the position on the $x$-axis immediately after the fourth step is $k = x-(4-x)=2x-4$, and
ii) the only possible positions of the particle immediately after the fourth step are
$k=-4,-2,0,2$ or $4$ as $x$ is $0, 1,2,3$, or $4$, respectively.
Hence, if we let $X$ denote the number of R steps taken by the particle in its four-step motion, in view of Eq.1, i) and ii), the pdf of $X$ is
$\begin{align*}
p_X(k) &= P(X=k),\ k=-4,-2,0,2,4 \\
&= P(X=2x-4),\ k=-4,-2,0,2,4 \\
&= P(Y=x),\ x=0,1,2,3,4 \\
\color{blue}{p_X(2x-4)} &\color{blue}{= \displaystyle {4\choose x}(2/3)^x(1/3)^{4-x},\ x=0,1,2,3,4}
\end{align*}$
