Suppose in a graph $G$ there is one pebble on each vertex except one vertex $v$ with $p(v) = 2$, then above pebbling problem has solution on $G \space iff \space G$ has a Hamiltonian circuit. It's easy to check if there is a Hamiltonian circuit, then there is a solution for pebbling on $G$.
On the other hand, in any solution to the pebbling, we should start from vertex $v$. Suppose that we visit some vertex $u$ twice such that this $u$ is the first vertex which visited twice in $G$ by pebbling algorithm, then we have a loop which starts from $u$ and ends in $u$ and finally because $u$ is the first for making loop then we have $p(u) = 1$ so we cannot continue pebbling algorithm. Indeed if the algorithm has a solution then we have $u=v$ which means we found a Hamiltonian circuit which starts in $v$.