Pd$\ce{Pd}$ can dissociate H2$\ce{H2}$ because the resulting Pd-H$\ce{Pd-H}$ bonds are more stable than the starting H2$\ce{H2}$. But the reason why Pd$\ce{Pd}$ is so good at dissociating H2$\ce{H2}$ is related to the energy barrier to bond formation. The dissociation of H2$\ce{H2}$ on a Pd$\ce{Pd}$ surface (and on Pt$\ce{Pt}$ and maybe several other metals) has no barrier. So you don't need to put the H2$\ce{H2}$ in an excited state to go over a barrier and create Pd-H$\ce{Pd-H}$ bonds. On Cu$\ce{Cu}$ for example, the bonding is possible but there is a barrier, you need to excite H2$\ce{H2}$ to dissociate it.
Here is a good theoretical articleHere is a good theoretical article (with a great title): http://web.mit.edu/andrew3/Public/Papers/1995/Hammer/1995_Nature_Why%20gold%20is%20the%20noblest_Hammer.pdf
It shows the different barriers to dissociation of H2$\ce{H2}$ on Pt$\ce{Pt}$, Ni$\ce{Ni}$, Cu$\ce{Cu}$ and Au$\ce{Au}$. It also gives an explanation for such differences (more physics than chemistry).