= Solution
Here <Cartesian category> means a category with all <finite limits>. Let $m:Y\hookrightarrow X$ be a <monomorphism>, and let $\varepsilon_X\hookrightarrow X\times PX$ be universal membership for a <power object> $PX$. Pull it back to $Y\times PX$ and then regard the result as a <subobject> of $X\times PX$. Its unique classifier is an endomorphism $r:PX\to PX$: on a parameterized subobject it intersects with $Y$. Intersecting twice has the same result as intersecting once, so uniqueness of classification gives $r^2=r$.
Take the <equalizer> $q:Q\hookrightarrow PX$ of $r$ and the identity, and restrict membership to $Y\times Q$. For any <subobject> $S\hookrightarrow Y\times Z$, its composite with $m\times1_Z$ is a <subobject> of $X\times Z$, hence is uniquely classified by $s:Z\to PX$. Since it is contained in $Y\times Z$, its intersection with that subobject changes nothing, so $rs=s$ and $s$ factors uniquely through $Q$. Conversely every map $Z\to Q$ gives the original kind of family by pulling back the restricted membership. These inverse operations give
$$
\boxed{\mathcal C(Z,Q)\cong\operatorname{Sub}(Y\times Z)},
$$
naturally in $Z$. Thus $Q$ is a <power object of a subobject>, even if no power objects other than $PX$ were assumed.
Now consider the <slice category> $\mathcal E/A$, which has <finite limits> formed by ambient pullbacks and has terminal object $1_A:A\to A$. The constant family $A\times X\to A$ has slice <power object> $A\times PX\to A$: for $Z\to A$,
$$
(A\times X)\times_A Z\cong X\times Z,
$$
and a classifier $Z\to PX$ is equivalently the map $Z\to A\times PX$ over $A$. Every family $p:X\to A$ is a <subobject> of this constant family through its monic graph $(p,1_X):X\hookrightarrow A\times X$. Apply the first construction inside the slice. \b[Consequently every slice of a Cartesian category with power objects again has power objects.]
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