Solution:
If $P$ is the incenter of $A B C$, then $\angle B P D = \angle A B P + \angle B A P = \frac{\hat{A} + \hat{B}}{2}$, and $\angle B D P = \angle B C P = \frac{\hat{C}}{2}$. From triangle $B D P$, it follows that $\angle P B D = 90^{\circ}$, i.e. that $E B$ is one of the altitudes of the triangle $D E F$. Similarly, $A D$ and $C F$ are altitudes, which means that $P$ is the orthocenter of $D E F$.

Notice that $A P$ separates $B$ from $C$, $B$ from $E$ and $C$ from $F$. Therefore $A P$ separates $E$ from $F$, which means that $P$ belongs to the interior of $\angle E D F$. It follows that $P \in \operatorname{Int}(\triangle D E F)$.
If $P$ is the orthocenter of $D E F$, then clearly $D E F$ must be acute. Let $A^{\prime} \in E F$, $B^{\prime} \in D F$ and $C^{\prime} \in D E$ be the feet of the altitudes. Then the quadrilaterals $B^{\prime} P A^{\prime} F$, $C^{\prime} P B^{\prime} D$, and $A^{\prime} P C^{\prime} E$ are cyclic, which means that $\angle B^{\prime} F A^{\prime} = 180^{\circ} - \angle B^{\prime} P A^{\prime} = 180^{\circ} - \angle B P A = \angle B F A$. Similarly, one obtains that $\angle C^{\prime} D B^{\prime} = \angle C D B$, and $\angle A^{\prime} E C^{\prime} = \angle A E C$.
- If $B \in \operatorname{Ext}(\triangle F P D)$, then $A \in \operatorname{Int}(\triangle E P F)$, $C \in \operatorname{Ext}(\triangle D P E)$, and thus $B \in \operatorname{Int}(\triangle F P D)$, contradiction.
- If $B \in \operatorname{Int}(\triangle F P D)$, then $A \in \operatorname{Ext}(\triangle E P F)$, $C \in \operatorname{Int}(\triangle D P E)$, and thus $B \in \operatorname{Ext}(\triangle F P D)$, contradiction.
This leaves us with $B \in F D$. Then we must have $A \in E F$, $C \in D E$, which means that $A = A^{\prime}$, $B = B^{\prime}$, $C = C^{\prime}$. Thus $A B C$ is the orthic triangle of triangle $D E F$ and it is well known that the orthocenter of an acute triangle $D E F$ is the incenter of its orthic triangle.