To solve this geometry problem involving triangle \( \Delta ABC \) with points \( M, N, P \) on segments \( AB, BC, \) and \( CA \), we follow the conditions provided:

1. **Given conditions**:
- \( \overline{BN} = \overline{BC} \)
- \( \overline{AQ} = 3 \overline{AC} \)

2. **Tasks**:
- (a) Find \( \overline{AN}, \overline{MB} \) in terms of \( \overline{AB}, \overline{AC} \).
- (b) Show that \( \overline{AN} \cdot 16 \overline{AT} = 9 \overline{AM} \cdot \overline{MR} \), where \( T \) and \( R \) are points of intersection or specific segments within the triangle.

### Solution Steps:

**(a) Finding \( \overline{AN} \) and \( \overline{MB} \)**:
- From the condition \( \overline{BN} = \overline{BC} \), we can deduce that \( N \) divides the segment \( BC \) such that \( N \) is actually at point \( C \).
- Using \( A \) as a reference point:
- \( \overline{AN} = \overline{AC} \)
- For \( M \) on segment \( AB \), assume \( \overline{AB} = x \) and point \( M \) divides it in some ratio, say \( k : 1 \). Then:
- \( \overline{MB} = \frac{x}{k + 1} \)

**(b) Proving \( \overline{AN} \cdot 16 \overline{AT} = 9 \overline{AM} \cdot \overline{MR} \)**:
- If specific values are involved for \( AN, AT, AM, MR \), we could substitute them in and show equality based on the previously established relations from the segments.
- This might involve finding \( \overline{AT} \) based on point positions within the triangle using similar triangles or segment ratios.

### Final Remarks:
Make sure to visualize the triangle and the points properly for accurate computation. Employ coordinate geometry or vector analysis if necessary to simplify the algebra involved.