Answer
$\color{blue}{p_X(2x-4) = \displaystyle {4\choose x}1/16,\ x=0,1,2,3,4}$
Work Step by Step
Note that insofar as going right (R) or left (L) is concerned (i.e., without regard to the particle's step number), the particle's motion may be modeled by a binomial distribution where step R is a success and step L is a failure, each with probability $1/2$, the number of trials is $n=4$, and the binomial random variable, say $Y$, is the number of R steps taken. The pdf of $Y$ is
$\begin{align*}
p_Y(x) &= {4\choose k}(1/2)^x(1/2)^{4-x},\ x =0,1,2,3,4 \\
&= {4\choose x}(1/2)^4 \\
p_X(x) &= {4\choose x}1/16,\ x=0,1,2,3,4 \qquad \text{(Eq. 1)}
\end{align*}$.
In addition, the particle's position immediately after the fourth step (or fourth-step position) on the $x$-axis may be computed as the number of R steps minus the number of L steps. Thus, if the particle made $x$ R steps out of four steps taken so that it must have made $4-x$ L steps, then the position on the $x$-axis of the particle must be at $\#({\rm R}) - \#({\rm L}) = x-(4-x) = 2x-4$.
Hence, if we let $X$ denote the fourth-step position of the particle on the $x$-axis, then $X = 4,2,0,-2$, or $-4$ as $X=2x-4$ for $x=0,1,2,3,4$, $x$ being the number of R steps taken by the particle in its four-step motion.
Thus, owing to the pdf of $Y$ in Eq. 1 based on $x$, the number of R steps taken by the particle, the pdf of $X$ is
$\begin{align*}
p_X(2x-4) &= P(X=2x-4),\ x=0,1,2,3,4 \\
&= P(Y=x),\ x=0,1,2,3,4 \\
\color{blue}{p_X(2x-4)} &\color{blue}{= \displaystyle {4\choose x}1/16,\ x=0,1,2,3,4}
\end{align*}$
