The answer for $\bigwedge_t$ is no. Perhaps the idea here can be adapted to $\bigvee_t$.

Consider the propositional formula $t \vee (t \to x)$ in the complete Heyting algebra of open subsets of $\mathbb{R}$.

Claim. For any open set $U \subseteq \mathbb{R}$, $\bigwedge_t t \vee (t \to U)$ is the set $U^\bullet := \{r \in \mathbb{R} : r \in U\text{ or }r\text{ is isolated in }\mathbb{R} \setminus U\}$.

Proof of claim. Clearly if $r \in U$, then $r \in (t \vee (t \to U))$ for any open set $t$. If $r$ is isolated in $\mathbb{R}\setminus U$, then for any open set $t$, we either have that $r \in t$ or $(\mathbb{R} \setminus t) \cup U$ contains a neigbhorhood of $r$. Conversely, if $r\notin U$ but $r$ is also not isolated in $\mathbb{R} \setminus U$, then $r \notin ((\mathbb{R} \setminus \{r\}) \vee ((\mathbb{R} \setminus \{r\}) \to U))$.

So now consider some propositional formula $\varphi(x,\bar{p})$. We need to show that this fails to be equal to $\bigwedge_{t} t \vee(t \to x)$ for all open sets $x$. Since there are only finitely many open sets in the tuple $\bar{p}$, we can find a non-empty open set $U \subseteq \mathbb{R}$ such that for each $i < |\bar{p}|$, either $U\wedge p_i = U$ or $U \wedge p_i = \bot$. For each $i < |\bar{p}|$, let $q_i = \top$ if $U \wedge p_i = U$ and let $q_i = \bot$ if $U \wedge p_i = \bot$. We now have that for any open $V$, $U \wedge \varphi(V,\bar{p}) = U \wedge \varphi(V,\bar{q})$.

Let $F$ be a closed subset of $U$ that is homeomorphic to Cantor space plus a single isolated point. Let $r$ be the single isolated point of $F$. Let $V = \mathbb{R} \setminus F$. Clearly we have that $V^\bullet = V \cup \{r\}$.

Claim. $U\wedge \varphi(V,\bar{q})$ is either $U$, $U \wedge V$, or $\bot$.

Proof of claim. We prove this by induction on propositional formulas in the single variable $x$ (i.e., we prove that for any propositional formula $\psi(x)$, $U \wedge \psi(V) \in \{U,U \wedge V,\bot\}$). Clearly we have that the statement is true for $U \wedge \bot = \bot$, $U \wedge \top = U$, and $U \wedge x$ (which is $U \wedge V$ when $x=V$).

If the statement is true for two propositional formulas $\psi(x)$ and $\chi(x)$, then it's easy to check that the statement is true for $\psi(x) \wedge \chi(x)$ and $\psi(x) \vee \chi(x)$. This just leaves $\psi(x) \to \chi(x)$. If $U \wedge \psi(V) = \bot$ or $U \wedge \chi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = U \in \{U,U \wedge V,\bot\}$. If $U \wedge \psi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = \chi(V) \in \{U,U \wedge V,\bot\}$. Finally, if $U \wedge \psi(V) = U \wedge V$ and $U \wedge \chi(V) $ is $U$ or $U \wedge V$, then $U \wedge ( \psi(V) \to \chi(V)) = U \wedge V \in \{U,U \wedge V,\bot\}$. So in every case, we have that $U \wedge (\psi(V) \to \chi(V)) \in \{U,U \wedge V, \bot\}$. Therefore, by induction, the same is true for $U \wedge \varphi(V,\bar{q})$.

Finally, note that $U \cap V^\bullet \notin \{U,U\wedge V, \bot\}$, whence $\varphi(V,\bar{p}) \neq V^\bullet = \bigwedge_t t \vee (t \to V)$.

Since we can do this for any formula $\varphi(x,\bar{p})$, we have that $\bigwedge_t t \vee (t \to V)$ is not equal to any propositional formula with parameters.

The answer for $\bigwedge_t$ is no. Perhaps the idea here can be adapted to $\bigvee_t$.

Consider the propositional formula $t \vee (t \to x)$ in the complete Heyting algebra of open subsets of $\mathbb{R}$.

Claim. For any open set $U \subseteq \mathbb{R}$, $\bigwedge_t t \vee (t \to U)$ is the set $U^\bullet := \{r \in \mathbb{R} : r \in U\text{ or }r\text{ is isolated in }\mathbb{R} \setminus U\}$.

Proof of claim. Clearly if $r \in U$, then $r \in (t \vee (t \to U))$ for any open set $t$. If $r$ is isolated in $\mathbb{R}\setminus U$, then for any open set $t$, we either have that $r \in t$ or $(\mathbb{R} \setminus t) \cup U$ contains a neigbhorhood of $r$. Conversely, if $r\notin U$ but $r$ is also not isolated in $\mathbb{R} \setminus U$, then $r \notin ((\mathbb{R} \setminus \{r\}) \vee ((\mathbb{R} \setminus \{r\}) \to U))$.

So now consider some propositional formula $\varphi(x,\bar{p})$. We need to show that this fails to be equal to $\bigwedge_{t} t \vee(t \to x)$ for all open sets $x$. Since there are only finitely many open sets in the tuple $\bar{p}$, we can find a non-empty open set $U \subseteq \mathbb{R}$ such that for each $i < |\bar{p}|$, either $U\wedge p_i = U$ or $U \wedge p_i = \bot$. For each $i < |\bar{p}|$, let $q_i = \top$ if $U \wedge p_i = U$ and let $q_i = \bot$ if $U \wedge p_i = \bot$. We now have that for any open $V$, $U \wedge \varphi(V,\bar{p}) = U \wedge \varphi(V,\bar{q})$.

Let $F$ be a closed subset of $U$ that is homeomorphic to Cantor space plus a single isolated point. Let $r$ be the single isolated point of $F$. Let $V = \mathbb{R} \setminus F$. Clearly we have that $V^\bullet = V \cup \{r\}$.

Claim. $U\wedge \varphi(V,\bar{q})$ is either $U$, $U \wedge V$, or $\bot$.

Proof of claim. We prove this by induction on propositional formulas in the single variable $x$ (i.e., we prove that for any propositional formula $\psi(x)$, $U \wedge \psi(V) \in \{U,U \wedge V,\bot\}$). Clearly we have that the statement is true for $U \wedge \bot = \bot$, $U \wedge \top = U$, and $U \wedge x$ (which is $U \wedge V$ when $x=V$).

If the statement is true for two propositional formulas $\psi(x)$ and $\chi(x)$, then it's easy to check that the statement is true for $\psi(x) \wedge \chi(x)$ and $\psi(x) \vee \chi(x)$. This just leaves $\psi(x) \to \chi(x)$. If $U \wedge \psi(V) = \bot$ or $U \wedge \chi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = U \in \{U,U \wedge V,\bot\}$. If $U \wedge \psi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = \chi(V) \in \{U,U \wedge V,\bot\}$. Finally, if $U \wedge \psi(V) = U \wedge V$ and $U \wedge \chi(V) $ is $U$ or $U \wedge V$, then $U \wedge ( \psi(V) \to \chi(V)) = U \wedge V \in \{U,U \wedge V,\bot\}$. So in every case, we have that $U \wedge (\psi(V) \to \chi(V)) \in \{U,U \wedge V, \bot\}$. Therefore, by induction, the same is true for $U \wedge \varphi(V,\bar{q})$.

Finally, note that $U \wedge V^\bullet \notin \{U,U\wedge V, \bot\}$, whence $\varphi(V,\bar{p}) \neq V^\bullet = \bigwedge_t t \vee (t \to V)$.

Since we can do this for any formula $\varphi(x,\bar{p})$, we have that $\bigwedge_t t \vee (t \to V)$ is not equal to any propositional formula with parameters.

The answer for $\bigwedge_t$ is no. Perhaps the idea here can be adapted to $\bigvee_t$.

Consider the propositional formula $t \vee (t \to x)$ in the complete Heyting algebra of open subsets of $\mathbb{R}$.

Claim. For any open set $U \subseteq \mathbb{R}$, $\bigwedge_t t \vee (t \to U)$ is the set $U^\bullet := \{r \in \mathbb{R} : r \in U\text{ or }r\text{ is isolated in }\mathbb{R} \setminus U\}$.

Proof of claim. Clearly if $r \in U$, then $r \in (t \vee (t \to U))$ for any open set $t$. If $r$ is isolated in $\mathbb{R}\setminus U$, then for any open set $t$, we either have that $r \in t$ or $(\mathbb{R} \setminus t) \cup U$ contains a neigbhorhood of $r$. Conversely, if $r\notin U$ but $r$ is also not isolated in $\mathbb{R} \setminus U$, then $r \notin ((\mathbb{R} \setminus \{r\}) \vee ((\mathbb{R} \setminus \{r\}) \to U))$.

So now consider some propositional formula $\varphi(x,\bar{p})$. We need to show that this fails to be equal to $\bigwedge_{t} t \vee(t \to x)$ for any open set $x$. Since there are only finitely many open sets in the tuple $\bar{p}$, we can find a non-empty open set $U \subseteq \mathbb{R}$ such that for each $i < |\bar{p}|$, either $U\wedge p_i = U$ or $U \wedge p_i = \bot$. For each $i < |\bar{p}|$, let $q_i = \top$ if $U \wedge p_i = U$ and let $q_i = \bot$ if $U \wedge p_i = \bot$. We now have that for any open $V$, $U \wedge \varphi(V,\bar{p}) = U \wedge \varphi(V,\bar{q})$.

Let $F$ be a closed subset of $U$ that is homeomorphic to Cantor space plus a single isolated point. Let $r$ be the single isolated point of $F$. Let $V = \mathbb{R} \setminus F$. Clearly we have that $V^\bullet = V \cup \{r\}$.

Claim. $U\wedge \varphi(V,\bar{q})$ is either $U$, $U \wedge V$, or $\bot$.

Proof of claim. We prove this by induction on propositional formulas in the single variable $x$ (i.e., we prove that for any propositional formula $\psi(x)$, $U \wedge \psi(V) \in \{U,U \wedge V,\bot\}$). Clearly we have that the statement is true for $U \wedge \bot = \bot$, $U \wedge \top = U$, and $U \wedge x$ (which is $U \wedge V$ when $x=V$).

If the statement is true for two propositional formulas $\psi(x)$ and $\chi(x)$, then it's easy to check that the statement is true for $\psi(x) \wedge \chi(x)$ and $\psi(x) \vee \chi(x)$. This just leaves $\psi(x) \to \chi(x)$. If $U \wedge \psi(V) = \bot$ or $U \wedge \chi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = U \in \{U,U \wedge V,\bot\}$. If $U \wedge \psi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = \chi(V) \in \{U,U \wedge V,\bot\}$. Finally, if $U \wedge \psi(V) = U \wedge V$ and $U \wedge \chi(V) $ is $U$ or $U \wedge V$, then $U \wedge ( \psi(V) \to \chi(V)) = U \wedge V \in \{U,U \wedge V,\bot\}$. So in every case, we have that $U \wedge (\psi(V) \to \chi(V)) \in \{U,U \wedge V, \bot\}$. Therefore, by induction, the same is true for $U \wedge \varphi(V,\bar{q})$.

Finally, note that $U \cap V^\bullet \notin \{U,U\wedge V, \bot\}$, whence $\varphi(V,\bar{p}) \neq V^\bullet = \bigwedge_t t \vee (t \to V)$.

Since we can do this for any formula $\varphi(x,\bar{p})$, we have that $\bigwedge_t t \vee (t \to V)$ is not equal to any propositional formula with parameters.

The answer for $\bigwedge_t$ is no. Perhaps the idea here can be adapted to $\bigvee_t$.

Consider the propositional formula $t \vee (t \to x)$ in the complete Heyting algebra of open subsets of $\mathbb{R}$.

Claim. For any open set $U \subseteq \mathbb{R}$, $\bigwedge_t t \vee (t \to U)$ is the set $U^\bullet := \{r \in \mathbb{R} : r \in U\text{ or }r\text{ is isolated in }\mathbb{R} \setminus U\}$.

Proof of claim. Clearly if $r \in U$, then $r \in (t \vee (t \to U))$ for any open set $t$. If $r$ is isolated in $\mathbb{R}\setminus U$, then for any open set $t$, we either have that $r \in t$ or $(\mathbb{R} \setminus t) \cup U$ contains a neigbhorhood of $r$. Conversely, if $r\notin U$ but $r$ is also not isolated in $\mathbb{R} \setminus U$, then $r \notin ((\mathbb{R} \setminus \{r\}) \vee ((\mathbb{R} \setminus \{r\}) \to U))$.

So now consider some propositional formula $\varphi(x,\bar{p})$. We need to show that this fails to be equal to $\bigwedge_{t} t \vee(t \to x)$ for all open sets $x$. Since there are only finitely many open sets in the tuple $\bar{p}$, we can find a non-empty open set $U \subseteq \mathbb{R}$ such that for each $i < |\bar{p}|$, either $U\wedge p_i = U$ or $U \wedge p_i = \bot$. For each $i < |\bar{p}|$, let $q_i = \top$ if $U \wedge p_i = U$ and let $q_i = \bot$ if $U \wedge p_i = \bot$. We now have that for any open $V$, $U \wedge \varphi(V,\bar{p}) = U \wedge \varphi(V,\bar{q})$.

Let $F$ be a closed subset of $U$ that is homeomorphic to Cantor space plus a single isolated point. Let $r$ be the single isolated point of $F$. Let $V = \mathbb{R} \setminus F$. Clearly we have that $V^\bullet = V \cup \{r\}$.

Claim. $U\wedge \varphi(V,\bar{q})$ is either $U$, $U \wedge V$, or $\bot$.

Proof of claim. We prove this by induction on propositional formulas in the single variable $x$ (i.e., we prove that for any propositional formula $\psi(x)$, $U \wedge \psi(V) \in \{U,U \wedge V,\bot\}$). Clearly we have that the statement is true for $U \wedge \bot = \bot$, $U \wedge \top = U$, and $U \wedge x$ (which is $U \wedge V$ when $x=V$).

If the statement is true for two propositional formulas $\psi(x)$ and $\chi(x)$, then it's easy to check that the statement is true for $\psi(x) \wedge \chi(x)$ and $\psi(x) \vee \chi(x)$. This just leaves $\psi(x) \to \chi(x)$. If $U \wedge \psi(V) = \bot$ or $U \wedge \chi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = U \in \{U,U \wedge V,\bot\}$. If $U \wedge \psi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = \chi(V) \in \{U,U \wedge V,\bot\}$. Finally, if $U \wedge \psi(V) = U \wedge V$ and $U \wedge \chi(V) $ is $U$ or $U \wedge V$, then $U \wedge ( \psi(V) \to \chi(V)) = U \wedge V \in \{U,U \wedge V,\bot\}$. So in every case, we have that $U \wedge (\psi(V) \to \chi(V)) \in \{U,U \wedge V, \bot\}$. Therefore, by induction, the same is true for $U \wedge \varphi(V,\bar{q})$.

Finally, note that $U \cap V^\bullet \notin \{U,U\wedge V, \bot\}$, whence $\varphi(V,\bar{p}) \neq V^\bullet = \bigwedge_t t \vee (t \to V)$.

Since we can do this for any formula $\varphi(x,\bar{p})$, we have that $\bigwedge_t t \vee (t \to V)$ is not equal to any propositional formula with parameters.