Answer: 351

Solution:

Group the vertices into classes $\mathcal{A} = \{S, P_1\},$ $\mathcal{B} = \{P_2, P_5\},$ and $\mathcal{C} = \{P_3, P_4\}.$ Each vertex of $\mathcal{A}$ adjoins one vertex of $\mathcal{A}$ and one of $\mathcal{B};$ each vertex of $\mathcal{B}$ adjoins one of $\mathcal{A}$ and one of $\mathcal{C};$ and each vertex of $\mathcal{C}$ adjoins one of $\mathcal{B}$ and the absorbing vertex $E.$ Hence if $a_n,$ $b_n,$ $c_n$ count the $n$-jump paths from $S$ that have not yet reached $E$ and end in each class, $$\begin{aligned} &a_{n+1} = a_n + b_n, \\ &b_{n+1} = a_n + c_n, \\ &c_{n+1} = b_n, \end{aligned}$$ and exactly $c_n$ paths reach $E$ for the first time on jump $n + 1.$

Starting from $(a_0, b_0, c_0) = (1, 0, 0),$ the values of $c_n$ for $n = 0, 1, \ldots, 11$ are $0,$ $0,$ $1,$ $1,$ $3,$ $4,$ $9,$ $14,$ $28,$ $47,$ $89,$ $155.$

The number of sequences of at most $12$ jumps ending at $E$ is $c_0 + c_1 + \cdots + c_{11} = 351.$