I haven't checked but here are some guidelines to show the group is Hopfian.

Write the original group (given by $4\times 4$ triangular matrices) as $G=D\ltimes U$ with $D=Z^2$ and $U$ its unipotent part. Set $U^2=[U,U]$ and $U^3=[U,U^2]$, which is central and naturally isomorphic to $F_p[t,1/t]$. Our group is $H=G/M$, where $M\subset U^3$ is generated by $F_p[t]$ and $t^{-2}$. Let $f$ be a surjective endomorphism of $H$.

1) check that the center of $G$ is precisely $U^3$.
It follows that $f$ induces a surjective endomorphism of $G/U^3$. Since this group is linear, it is Hopfian so this is an automorphism of $G/U^3$.

2) Describe the group of automorphisms of $G/U^2 = Z^2\ltimes F_p[t,1/t]^3$. (It should be reasonably easy to describe). 

3) Deduce a description of the group of automorphisms of $G/U^3$, or at least describe how these automorphisms act on $U^2/U^3$, showing that modulo something, the coefficient $12$ is multiplied by a monomial $w\cdot t^a$ ($w\in F_p*$) and the coefficient $24$ is multiplied by $vt^b$. So, taking a commutator (that should kill the "modulo something"), we obtain that in the action of $f$ on $H$, the coefficient $14$ should be multiplied by a monomial. This multiplication should stabilize $M$ so this is multiplication by a scalar in $F_p*$, which implies that f is actually an automorphism.