I agree with David that a bit more background/explanation would help. For instance, I'm not sure what you mean by a "big vector bundle". *Let me suppose that $E$ is a big line bundle, i.e. a line bundle with positive self-intersection.
Then the answer is no.*

To see this, note that $E$ is nef i.e. $E\cdot C\ge 0$ for any curve
$C$, since $C$ can be moved into general position without affecting the intersection number
because $A$ is homogenous. It follows that $E\otimes P$ is nef and big for any $P\in Pic^0(A)$ because $E\otimes P$ and $E$ are numerically equivalent. Therefore $H^i(E\otimes P)=0$ for $i>0$ by Kawamata-Viehweg vanishing. If this held for $i=0$, then the euler characteristic
$\chi(E\otimes P)=\chi(E)=0$. On an abelian variety, Riemann-Roch gives $\chi(E) = E^n/n!$
where $n=\dim A$ (see Mumford's Abelian varieties). Therefore $E^n=0$ contradicting bigness.