In a quadrilateral $ABCD$ sides $AB$ and $CD$ are equal, $\angle A=150^\circ,$ $\angle B=44^\circ,$ $\angle C=72^\circ.$ 
Perpendicular bisector of the segment $AD$ meets the side $BC$ at point $P.$ 
Find $\angle APD.$

[i]Proposed by F. Bakharev[/i]
1. **Identify the given information and setup the problem:**
   - Quadrilateral \(ABCD\) with \(AB = CD\).
   - Angles: \(\angle A = 150^\circ\), \(\angle B = 44^\circ\), \(\angle C = 72^\circ\).
   - Perpendicular bisector of segment \(AD\) meets side \(BC\) at point \(P\).

2. **Construct auxiliary points and angles:**
   - Let \(A'\) be a point on \(BC\) such that \(AA' = AB\). Then \(\angle BAA' = 180^\circ - 2 \cdot 44^\circ = 92^\circ\).
   - Similarly, let \(D'\) be a point on \(BC\) such that \(DD' = CD\). Then \(\angle CDD' = 180^\circ - 2 \cdot 72^\circ = 36^\circ\).

3. **Find the intersection point \(E\) of \(AA'\) and \(DD'\):**
   - We have \(\angle EAD = 150^\circ - 92^\circ = 58^\circ\).
   - We also have \(\angle EDA = 94^\circ - 36^\circ = 58^\circ\).
   - Therefore, triangle \(EAD\) is isosceles, implying \(EA = ED\).

4. **Prove that \(E\) lies on \(BC\):**
   - Assume \(E \not\in BC\).
   - The line \(BC\) bisects the plane into two half-planes.
   - If \(E\) is on the same half-plane as \(A\) and \(D\), then \(EA' = AA' - EA\) and \(ED' = DD' - ED\).
   - If \(E\) is on the opposite half-plane, then \(EA' = EA - AA'\) and \(ED' = ED - DD'\).
   - In both cases, from \(EA = ED\) and \(AA' = DD'\), we get \(EA' = ED'\), making \(\triangle EA'D'\) isosceles.
   - This is a contradiction because the lines \(AA'\) and \(DD'\) make different angles with line \(BC\) (\(44^\circ\) and \(72^\circ\)).

5. **Conclude that \(E\) must lie on \(BC\):**
   - Since assuming \(E \not\in BC\) leads to a contradiction, \(E\) must lie on \(BC\).
   - Therefore, \(E\) is the point \(P\).

6. **Determine \(\angle APD\):**
   - Since \(E\) is on \(BC\), the segments \(EA'\) and \(ED'\) are equal, but they have zero length.
   - Thus, \(\angle APD = 180^\circ - (\angle B + \angle C) = 180^\circ - (44^\circ + 72^\circ) = 64^\circ\).

The final answer is \(\boxed{64^\circ}\)