Suppose first that $p\Vdash^*\varphi$. If $G$ is any <generic filter> over $M$ containing $p$, the <forcing theorem> gives $M[G]\models\varphi$. Hence $p\Vdash\varphi$ in the <semantic forcing relation>.

Conversely, suppose $p\not\Vdash^*\varphi$. By the stated density equivalence for the <syntactic forcing relation>, some $r\leq p$ satisfies $r\Vdash^*\neg\varphi$. The <Rasiowa–Sikorski lemma> supplies a generic filter $G$ over the countable model $M$ containing $r$, and upward closure then gives $p\in G$. The <forcing theorem> yields $M[G]\models\neg\varphi$, so $p\not\Vdash\varphi$ semantically. Therefore
$$
\boxed{p\Vdash\varphi\quad\Longleftrightarrow\quad p\Vdash^*\varphi.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 121 / 3 / iii / Solution

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If $M$ is a <countable transitive model>, $\mathbb P\in M$, and $p\in\mathbb P$, enumerate the dense subsets of $\mathbb P$ belonging to $M$ and recursively choose a decreasing sequence starting below $p$ that meets each one. Its upward closure is a <generic filter> over $M$ containing $p$.

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 Incoming links: Rasiowa–Sikorski lemma