Problem 880

We denote the smallest angle required to rotate line $l$ counterclockwise to be parallel to line $n$ as $\angle(l, n)$.

Lemma: Four non-collinear points $P, Q, R, S$ are concyclic if and only if $\angle(Q P, Q R)=\angle(S P, S R)$ (Figure 1). This can be proven by the properties of angles subtended by the same arc in a circle.

Construct $A K B C(K \in S)$ (Figure 2), and extend $K D$ to intersect circle $S$ at $P_{0}$. We will prove that for every point $M$ satisfying the problem's conditions, $P$ coincides with $P_{0}$. There are two possible cases.
(1) Point $N$ and $P_{0}$ do not coincide. Since points $A, K, P_{0}$, and $N$ are all on circle $S$, we have $\angle\left(A K, K P_{0}\right)=\angle(A N, N P_{0})$. Since $B C A K$, then $\angle\left(A K, K P_{0}\right)=\angle\left(B C, K P_{0}\right)$. This implies that $\angle\left(M D, D P_{0}\right)=\angle\left(M N, N P_{0}\right)$, meaning points $M, N, P_{0}, D$ are concyclic, thus $P_{0}=P_{0}$.
(2) Point $N$ and $P_{0}$ coincide. In a homothety centered at point $P_{0}$, point $K$ is moved to point $D$, and line $A P$ is transformed into itself, while line $A K$ is transformed into line $M D$ (since $A K \parallel M D$). This means point $A$ is moved to point $M$. Therefore, circle $S$ is transformed into the circumcircle of triangle $N M D$. Since the center of homothety is point $P_{0}=N$, these circles have no other common points, i.e., $P=P_{0}$.

Clearly, the point $M$ is the projection of point $P_{0}$ onto line $B C$, and this projection lies within segment $B C$ (since $\angle A$ is acute, and point $M$ does not coincide with point $D$, because $\angle K D C = \angle D K B + \angle K B D$ is an obtuse angle).