= Solution
Let $P:\mathcal E\to\mathcal D$ be a <discrete fibration> and let $m:A\to A'$ be monic in $\mathcal E$. If $u,v:B\to P(A)$ satisfy $P(m)u=P(m)v$, lift $u,v$ uniquely to $\widetilde u,\widetilde v$ with codomain $A$. The composites $m\widetilde u,m\widetilde v$ are lifts of the same arrow with codomain $A'$, so uniqueness gives equality. Since $m$ is monic, $\widetilde u=\widetilde v$, hence $u=v$. Thus $P(m)$ is monic.
Use the convention that $(F\downarrow B)$ has objects $(A,u:FA\to B)$. Its forgetful functor sends $(A,u)$ to $A$. Given $h:C\to A$, the unique arrow above $h$ with codomain $(A,u)$ has domain $(C,uFh)$. Hence the forgetful functor is a discrete fibration.
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