Let's consider two propositions $\phi$ and $\psi$ of a propositional language $L$.
Let's suppose that $\phi \vDash \psi$, that is, that for every Boolean algebra $B$ and for every evaluation function $V:\text{Prop}(L) \rightarrow B$ we have that if $V(\phi)=1$ then $V(\psi)=1$.
Can we prove that for every Boolean algebra $B$ and for every evaluation function $V:\text{Prop}(L) \rightarrow B$ we have that $V(\phi) \le V(\psi)$?
Yes. In fact, we prove something somewhat more general.
Let $S$ be a set of propositions over $L$, and let $\psi$ be a proposition over $L$. Suppose $S \models \psi$. That is, suppose that for all Boolean algebras $B$ and evaluation functions $V$, if for all $s \in S$, $V(s) = 1$, then $V(\psi) = 1$.
Then for all Boolean algebras $B$ and evaluation functions $V$ on $B$, there exists some $n$ and some $s_1, \ldots, s_n$ such that $V(s_1 \land \cdots \land s_n) \leq V(\psi)$.
Proof: let $F$ be the filter generated by $V(S)$. That is,
$$F = \{b \in B \mid \exists n \in \mathbb{N} \exists s_1, \ldots, s_n \in S, V(s_1 \land \cdots \land s_n) \leq b\}$$
Note that $F$ is in fact a filter - it is upward closed and closed under finite meets (we don’t necessarily have that $0 \notin F$). Moreover, $S \subseteq F$. Then let $C = B / F$. If you are unfamiliar with this notation, we define an equivalence relation $\sim$ on $B$ by $x \sim y$ when $\exists f \in F (x \land f = y \land f)$. Then $C$ is the Boolean algebra with underlying set $B / \sim$, such that the quotient map $\pi : B \to B / \sim$ is a Boolean algebra homomorphism.
It’s easy to see from the definitions that $\pi^{-1}(\{1\}) = F$. So for all $s \in S$, $\pi(V(s))= 1$. $\pi \circ V$ is an evaluation map, so $\pi(V(\psi)) = 1$. Therefore, $V(\psi) \in F$. This is exactly to say that there is some sequence $s_1, \ldots, s_n$ of elements of $S$ such that $V(s_1 \land \cdots \land s_n) \leq V(\psi)$. $\square$
Returning to your question, we can take $S = \{\phi\}$. Take some sequence $s_1, \ldots, s_n$ of elements of $S$ such that $V(s_1 \land \cdots \land s_n) \leq V(\psi)$. Then $V(\phi) \leq V(s_1) \land \cdots \land V(s_n) = V(s_1 \land \cdots \land s_n) \leq V(\psi)$.