I think that in your assumption, the supremum is actually attained.

Consider the set 
$$\hat E:=\{tp\\ :\\  t\geq0\\ , \quad p\in E\\ \}\cap\bar B(0,1;\ell^\alpha).$$
Since $E$ is convex, $\hat E$ is convex too (here $\bar B(0,1;\ell^\alpha)$ denotes the closed unit ball of the sequence space $\ell^\alpha$).

Moreover, we are going to show that the assumption that $E^'$ is closed in $\ell^1,$ implies that $\hat E$ is a closed bounded subset of the reflexive space $\ell^\alpha$, thus weakly compact. Indeed, let $u$ belong to the $\ell^\alpha$ norm closure of $\hat E.$ So, there exists a sequence $t_j\geq0,$ and a sequence $p_j\in E,$ such that $u_j:=t_j\\ p_j$ converges to $u$ in $\ell^\alpha.$ If $u=0$ then $u\in \hat E$ and there's nothing to prove; otherwise we have (for large $j$) that $p_j/ \|p_j\|_\alpha= u_j/ \|u_j \|_\alpha,$ which converges in $\ell^\alpha$ to $u/\|u\|_\alpha.$  Hence $p^'_j:=\big(p_j/\|p_j \|_\alpha\big)^\alpha$ converges in $\ell^1$ to $\big(u/\|u \|_\alpha\big)^\alpha, $ showing that the latter belongs to $E^'$, which is $\|\cdot\|_1$-closed. This implies that for some $p\in E,$ $u$ has the form $\frac{\|u\|_\alpha}{\|p\|_\alpha}\\ p,$ so is in $\hat E$.

Now consider $v:=\big(q/\|q\|_\alpha\big)^{\alpha-1}.$ It is a norm-one element of $\ell^{\alpha^'},$ and your optimization problem can be rewritten as
$$s:=\sup_{p\in E}\big(\frac{p}{\|p\|_\alpha}\cdot v   \big) =\sup_{u\in\hat E} (u\cdot v),$$
that is attained by the weak compactness of $\hat E.$