Positive forms and strongly positive forms are bidual This question comes from the Complex Analytic and Differential Geometry by Demailly. Let $V$ be a $n$ dimensional complex space. Consider the exterior algebra $\Lambda V^* = \oplus \Lambda^{(p,q)}V^*$. $u \in \Lambda^{(q,q)}V^*$ is called strongly positive if
$$u= \sum \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$
where $\gamma_s \ge 0$. $v \in \Lambda^{(p,p)}V^*$ where $p+q = n$ is called positive if $u\wedge v$ is positive $(n,n)$-form, i.e. $u\wedge v = \lambda i\mathrm{d}z_1 \wedge \mathrm{d}\bar{z}_1 \wedge \cdots \wedge i\mathrm{d}z_p \wedge \mathrm{d}\bar{z}_p$ where $\lambda$ is a real positive number, for all $u$ strongly positive.
The question comes whether $u$ is strongly positive if $u\wedge v$ is positive for all $v$ positive. In fact this is true in real case where bidual of a convex cone is its closure. This is also the fact that Demailly uses to prove the duality.

The answer to your question is absolutely right.
The sets of positive and strongly positive forms are closed convex cones, i.e. closed and stable under convex combinations. Also, the positive cone  (denoted by $\Gamma_P$) is dual to the strongly positive cone (denoted by $\Gamma_{SP}$)  via the pairing
$$\Lambda^{p,p} V^*\times\Lambda^{q,q} V^*\longrightarrow \mathbb C$$
$$\qquad\quad(u,v)\qquad\quad\,\,\,\,\longmapsto u\wedge v\,/\,\tau,$$
where $\tau$ is the $(n,n)$-form determining the canonical orientation of $V$.
That is, $u\in\Lambda^{p,p}V^*$ is positive iff $u\wedge v\ge 0$ for all strongly positive forms $v\in\Lambda^{q,q} V^*.$
Using some knowledges from functional analysis, we know that the bidual of an arbitrary convex cone is equal to its closure. So, one has the following:
$$\begin{aligned} & v\,\,\text{is strongly positive} \\\iff & v\in\Gamma_{SP}\\\iff &v\in\,(\Gamma_{SP})^{**}\\\iff &v\in\,(\Gamma_P)^*\\\iff &\text{For all positive forms} \,\,u, \text{we have }v\wedge u=u\wedge v\ge 0. \end{aligned}$$
