PDF of a Shifted Rectangular Pulse
A random process is a collection of random variables, one for each time instant.  Here, the $t$-th random variable is denoted by $X(t)$ where the assumption is that $t \in (-\infty,\infty)$ and what you call the PDF is actually a whole collection of PDFs, one for each of the (uncountably infinitely many) random variables $X(t)$.  Now, we are told for any given (fixed) value of $t$, say $t=0.313$, the random variable $X(t)$ (specifically $X(0.313)$ in our chosen example) is a function (namely a rect function here) of the random variable $T_0$.
$$X(t) = \operatorname{rect}\left(\frac{t-T_0}{T}\right); \qquad
X(0.313) = \operatorname{rect}\left(\frac{0.313-T_0}{T}\right)$$  Now, the rect function can only take on values $0$ or $1$ and so we immediately can say that regardless of the numerical value of $t$ (doesn't have to be
$0.313$) or whatever value the
random variable $T_0$ might have assumed, $X(t)$ can only take on values $0$ or $1$, that is,


  for each value of $t \in (-\infty,\infty)$, $X(t)$ is a Bernoulli random variable. 


As such, $X(t)$ does not have a PDF but has a PMF (probability mass function), unless of course you like to use impulses and define what are called discrete density functions to incorporate both notions into one.

The only question remaining is what is the parameter $p$ (the probability that $X(t)$ equals $1$) of this Bernoulli random variable,
and the value of $p$ depends on what value we assume for $t$. That is, the answer is not something like $p = 0.3$ but rather a multi-line equation like
$$p(t) = \begin{cases}\cdots &amp; \text{if}~ t\in \cdots,\\
\cdots &amp; \text{if}~t\in \cdots,\\
0, &amp;\text{otherwise.}
\end{cases}$$
To figure out the details, note that $\operatorname{rect}(\cdot)$ has value $1$ exactly when its argument has value $\leq \frac 12$. Thus, as a function of $t$, $\operatorname{rect}\left(\frac{t-T_0}{T}\right)$ has value $1$ exactly when $$T_0-\frac 12T \leq t \leq T_0+\frac 12T.$$
"Wait a minute", you splutter, "Didn't you say that $t$ had a fixed value
such as $0.313$? What's all this about as a function of $t$??? $t$ is not a variable; it is a fixed value" Well, we need to turn the thing around a bit. Notice that for a fixed value of $t$ (happy now?) $X(t)$ has value $1$ if and only if $T_0$ is such that
$$t - \frac 12T \leq T_0 \leq t+\frac12T,$$
that is, for the chosen fixed value of $t$, and keeping in mind that
$T_0 \in [-T, T]$
\begin{align}
P\{X(t) = 1\} = p(t) &amp;= P\left(t - \frac 12T \leq T_0 \leq t+\frac12T\right)\\
&amp;= P\left(T_0 \in \left(\left[t- \frac 12T, t+ \frac 12T\right]\cap
\big[-T,T\big]\right)\right)
\end{align}
That intersection on the right depends on what we have chosen $t$ to be,
and so we consider various cases. For brevity, denote the interval
$\left[t- \frac 12T, t+ \frac 12T\right]$ as $\mathcal A$ and the interval
$[-T.T]$ as $\mathcal B$


If $\color{blue}{t &lt; -\frac{3T}{2}}$, then $t+\frac 12T &lt; -T$ and so the right end of the interval $\mathcal A$ is smaller than the left end of $\mathcal B$
and so $\mathcal A \cap \mathcal B = \emptyset$. We get that $p(t) = 0$
for $t \leq -\frac{3T}{2}$. Similarly, for $t &gt; \frac{3T}{2}$, the
left end $t-\frac 12T$ exceeds $T$, the right end of $\mathcal B$. So


  $p(t) = 0$ if $|t| &gt; \frac{3T}{2}$.

If $\color{blue}{-\frac{3T}{2} \leq t &lt; -\frac T2}$, then the right end $t+\frac T2$
is in $[-T,0]$ but the left end $t-\frac T2$ is smaller than $-T$, that
is, $\mathcal A \cap \mathcal B =  \left[-T, t+\frac T2\right]$.  Since
$T_0 \sim U[-T,T]$, we get that
$p(t) = \frac{1}{2T}\times \left(t+\frac T2 - (-T)\right)
= \frac{1}{2T}\times \left(t+\frac {3T}{2}\right)$. Note that
$p(t)$ increases from $0$ at $t=\frac{3T}{2}$ to $\frac 12$ as
$t$ approaches $-\frac T2$.  Similarly, 
for $\frac T2 &lt; t \leq \frac {3T}{2}$, $p(t)$ decreases from
$\frac 12$ to $0$ linearly.  We have that


  $p(t) = \frac{1}{2T}\times \left(\frac {3T}{2} - |t|\right)$ if
  $\frac T2 &lt; t \leq \frac{3T}{2}$

If $\color{blue}{|t| \leq \frac{T}{2}}$, then both the left end
$t-\frac T2$ and the right end $t+\frac T2$ of $\mathcal A$ are in
$\mathcal B$, and so 


  $p(t) = \frac 12$ if $|t| \leq \frac{T}{2}$.



The process in question is not what is called a stationary process and it does not have a PDF (or PMF) but rather a whole bunch of different PDFs (or PMFs), one for each time instant $t$. Read the first part of 
this answer to learn a little more about these ideas.