The answer is obviously yes by a compactness argument:

Consider $\Delta^d$ as a (compact) topological space. For any $x \in \Delta^d$, there is a parallelotope $P_x$ containing $x$ in $P_x'$ where $P_x'$ is the interior of $P_x$ in $\Delta^d$. (Note that the interior is indeed taken in $\Delta^d$ and not in $\mathbb{R}^d$.)

Thus $P'_x$ form an open cover of $\Delta^d$ and there is a finite subcover $\{P'_{x_i}\}$. The required set of parallelotopes is then $\{P_{x_i}\}$. 

This reasoining does not say what are the required parallelotopes. But I believe that it is actually not very hard to find the required finite set (not too big) of them by choosing $x_i$ and $P_{x_i}$ suitably.