Let me try to prove that this question is equivalent to the previous, namely,

Theorem. Assume that $\mathbf{P}$ is a finite set and $\mathscr{S}$ is a family of subsets of $\mathbf{P}$ which is closed under taking over-sets. Then there exists a finite set $X$ and an injection $\varphi:\mathbf{P}\to 2^X$ such that $$ \mathscr{S}=\{\mathbf{S}\subset \mathbf{P}:\cup_{j\in \mathbf{S}}\varphi(j)=X\}. $$

Proof. For any set $\mathbf{S}\subset \mathbf{P}$ such that $\mathbf{S}\notin \mathscr{S}$ choose an element $x_{\mathbf{S}}$ which does not belong to all sets $\varphi(i),i\in \mathbf{S}$, and does belong to all $\varphi(j),j\notin \mathbf{S}$. Define $X=\sqcup_{\mathbf{S}} \{x_{\mathbf{S}}\}$, $\varphi$ is already defined. If $\mathbf{S}\notin \mathscr{S}$, then $\cup_{j\in \mathbf{S}}\varphi(j)\ne X$, because of the element $x_{\mathbf{S}}$. Now take $\mathbf{S}\in \mathscr{S}$. Fix any element $x_{\mathbf{T}}\in X$, where $\mathbf{T}\notin \mathscr{S}$. Since all over-sets of $\mathbf{S}$ belong to $\mathscr{S}$, we conclude that $\mathbf{T}$ is not an over-set of $\mathbf{S}$, i.e., there exists $j\in \mathbf{S}\setminus \mathbf{T}$. The set $\varphi(j)$ covers $x_{\mathbf{T}}$. Since the element $x_{\mathbf{T}}\in X$ was arbitrary, we conclude that $\cup_{j\in \mathbf{S}}\varphi(j)=X$.

Let me try to prove that the question about minimal elements is equivalent to the previous, namely:

Theorem. Assume that $\mathbf{P}$ is a finite set and $\mathscr{S}$ is a family of subsets of $\mathbf{P}$ which is closed under taking over-sets. Then there exists a finite set $X$ and an injection $\varphi:\mathbf{P}\to 2^X$ such that $$ \mathscr{S}=\{\mathbf{S}\subset \mathbf{P}:\cup_{j\in \mathbf{S}}\varphi(j)=X\}. $$

Proof. For any set $\mathbf{S}\subset \mathbf{P}$ such that $\mathbf{S}\notin \mathscr{S}$ choose an element $x_{\mathbf{S}}$ which does not belong to all sets $\varphi(i),i\in \mathbf{S}$, and does belong to all $\varphi(j),j\notin \mathbf{S}$. Define $X=\sqcup_{\mathbf{S}} \{x_{\mathbf{S}}\}$, $\varphi$ is already defined. If $\mathbf{S}\notin \mathscr{S}$, then $\cup_{j\in \mathbf{S}}\varphi(j)\ne X$, because of the element $x_{\mathbf{S}}$. Now take $\mathbf{S}\in \mathscr{S}$. Fix any element $x_{\mathbf{T}}\in X$, where $\mathbf{T}\notin \mathscr{S}$. Since all over-sets of $\mathbf{S}$ belong to $\mathscr{S}$, we conclude that $\mathbf{T}$ is not an over-set of $\mathbf{S}$, i.e., there exists $j\in \mathbf{S}\setminus \mathbf{T}$. The set $\varphi(j)$ covers $x_{\mathbf{T}}$. Since the element $x_{\mathbf{T}}\in X$ was arbitrary, we conclude that $\cup_{j\in \mathbf{S}}\varphi(j)=X$.

Let me try to prove that this question is equivalent to the previous, namely,

Theorem. Assume that $\mathbf{P}$ is a finite set and $\mathscr{S}$ is a family of subsets of $\mathbf{P}$ which is closed under taking over-sets. Then there exists a finite set $X$ and an injection $\varphi:\mathbf{P}\to 2^X$ such that $$ \mathscr{S}=\{\mathbf{S}\subset \mathbf{P}:\cup_{j\in \mathbf{S}}\varphi(j)=X\}. $$

Proof. For any set $\mathbf{S}\subset \mathbf{P}$ such that $\mathbf{S}\notin \mathscr{S}$ choose an element $x_{\mathbf{S}}$ which does not belong to all sets $\varphi(i),i\in \mathbf{S}$, and does belong to all $\varphi(j),j\notin \mathbf{S}$. Define $X=\sqcup_{\mathbf{S}} \{x_{\mathbf{S}}\}$, $\varphi$ is already defined. If $\mathbf{S}\in \mathscr{S}$, then $\cup_{j\in \mathbf{S}}\varphi(j)\ne X$, because of the element $x_{\mathbf{S}}$. Now take $\mathbf{S}\notin \mathscr{S}$. Fix any element $x_{\mathbf{T}}\in X$, where $\mathbf{T}\notin \mathscr{S}$. Since all over-sets of $\mathbf{S}$ belong to $\mathscr{S}$, we conclude that $\mathbf{T}$ is not an over-set of $\mathbf{S}$, i.e., there exists $j\in \mathbf{S}\setminus \mathbf{T}$. The set $\varphi(j)$ covers $x_{\mathbf{T}}$. Since the element $x_{\mathbf{T}}\in X$ was arbitrary, we conclude that $\cup_{j\in \mathbf{S}}\varphi(j)=X$.

Let me try to prove that this question is equivalent to the previous, namely,

Theorem. Assume that $\mathbf{P}$ is a finite set and $\mathscr{S}$ is a family of subsets of $\mathbf{P}$ which is closed under taking over-sets. Then there exists a finite set $X$ and an injection $\varphi:\mathbf{P}\to 2^X$ such that $$ \mathscr{S}=\{\mathbf{S}\subset \mathbf{P}:\cup_{j\in \mathbf{S}}\varphi(j)=X\}. $$

Proof. For any set $\mathbf{S}\subset \mathbf{P}$ such that $\mathbf{S}\notin \mathscr{S}$ choose an element $x_{\mathbf{S}}$ which does not belong to all sets $\varphi(i),i\in \mathbf{S}$, and does belong to all $\varphi(j),j\notin \mathbf{S}$. Define $X=\sqcup_{\mathbf{S}} \{x_{\mathbf{S}}\}$, $\varphi$ is already defined. If $\mathbf{S}\notin \mathscr{S}$, then $\cup_{j\in \mathbf{S}}\varphi(j)\ne X$, because of the element $x_{\mathbf{S}}$. Now take $\mathbf{S}\in \mathscr{S}$. Fix any element $x_{\mathbf{T}}\in X$, where $\mathbf{T}\notin \mathscr{S}$. Since all over-sets of $\mathbf{S}$ belong to $\mathscr{S}$, we conclude that $\mathbf{T}$ is not an over-set of $\mathbf{S}$, i.e., there exists $j\in \mathbf{S}\setminus \mathbf{T}$. The set $\varphi(j)$ covers $x_{\mathbf{T}}$. Since the element $x_{\mathbf{T}}\in X$ was arbitrary, we conclude that $\cup_{j\in \mathbf{S}}\varphi(j)=X$.