Given a Leontief model with $n \times n$ input-output matrix $B$, whose diagonal elements are positive and off-diagonal elements are non-positive. There is a single unit of labor available to this economy. Let $x$ be the vector of labor allocated to each of the $n$ sectors. A vector $y$ is a net output of the system provided $y=Bx$ for some $x\ge 0,$ and it is feasible whenever $\sum_{j=1}^n x_j \le 1$. A vector $\bar{y}$ is efficient if $\bar{y}$ is feasible and $y\ge \bar{y}, y \ne \bar{y}$ implies $y$ is not feasible. The set of efficient net outputs is denoted $E(B)$. Finally, a Leontief model is productive if there exists $y^*>>0, x^*\ge0$ such that $y^*=Bx^*.$
I would like to prove the following claim: Suppose the Leontief model is productive. Then, there is a price vector $p>>0$ such that $$E(B)=\{y\in\mathbb{R}_+^n:p \cdot y=1\}.$$
Here, the labor wage rate is the numeraire and normalized to 1.
The set of feasible outputs is the image $B(H)$ of the half-space $H = \{ x \in \mathbb{R}^n, \sum x_i \leq 1\} $ under the linear map $B$. $B(H)$ is a convex set and $E(B)$ is a face of it.
Productivity implies that $B(H)$ is full-dimensional and has compact intersection with $\mathbb{R}^n_+$, because an element $x^*$ in the standard simplex $\Delta_n$ is mapped to the interior of $\mathbb{R}^n_+$.
In other words, $B(H)$ is itself a half-space of the form $\{ y \in \mathbb{R}^n_+:\; p \cdot y \leq 1\}$. So you're done.