= Solution
For $X\subseteq A$, define
$$
\exists_p(X)=p[X],\qquad \forall_p(X)=\{b\in B:p^{-1}\{b\}\subseteq X\}=B\setminus p[A\setminus X].
$$
Both are monotone maps $PA\to PB$. For $Y\subseteq B$,
$$
p[X]\subseteq Y\quad\Longleftrightarrow\quad X\subseteq p^{-1}[Y],
$$
so <direct image> is a <left adjoint> to inverse image. Also,
$$
p^{-1}[Y]\subseteq X\quad\Longleftrightarrow\quad Y\subseteq\{b:p^{-1}\{b\}\subseteq X\},
$$
so the fiberwise universal condition gives the <right adjoint>. Therefore
$$
\boxed{\exists_p\dashv p^*\dashv\forall_p.}
$$
An empty fiber satisfies the universal condition vacuously, so every point outside $p[A]$ belongs to $\forall_p(X)$, regardless of $X$. This is essential when $p$ is not surjective.
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