Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.
1. **Identify the given information and setup the problem:**
   - Triangle \(ABC\) with \(AC = 450\) and \(BC = 300\).
   - Points \(K\) and \(L\) are on \(\overline{AC}\) and \(\overline{AB}\) respectively such that \(AK = CK\) and \(\overline{CL}\) is the angle bisector of \(\angle C\).
   - Point \(P\) is the intersection of \(\overline{BK}\) and \(\overline{CL}\).
   - Point \(M\) is on line \(\overline{BK}\) such that \(K\) is the midpoint of \(\overline{PM}\).
   - Given \(AM = 180\), find \(LP\).

2. **Prove that \(AMCP\) is a parallelogram:**
   - Since \(AK = CK\), point \(K\) is the midpoint of \(\overline{AC}\), so \(AK = CK = 225\).
   - Since \(K\) is the midpoint of \(\overline{PM}\), \(PM = 2 \cdot PK\).
   - Given \(AM = 180\), and since \(K\) is the midpoint of \(\overline{PM}\), \(PM = 2 \cdot 180 = 360\).
   - Since \(AK = CK\) and \(K\) is the midpoint of \(\overline{PM}\), \(AMCP\) forms a parallelogram because the diagonals bisect each other.

3. **Use Menelaus' Theorem on \(\triangle ALC\) with respect to \(\overline{BM}\):**
   - Menelaus' Theorem states that for a transversal intersecting the sides of a triangle, the product of the ratios of the segments is 1.
   - Applying Menelaus' Theorem to \(\triangle ALC\) with transversal \(\overline{BM}\):
     \[
     \frac{CQ}{QA} \cdot \frac{AB}{BL} \cdot \frac{LP}{PC} = 1
     \]
   - Since \(K\) is the midpoint of \(\overline{AC}\), \(CQ = QA = 225\).
   - Substituting \(CQ = QA = 225\) and \(PC = 180\):
     \[
     \frac{225}{225} \cdot \frac{AB}{BL} \cdot \frac{LP}{180} = 1
     \]

4. **Use the Angle Bisector Theorem to find \(\frac{AB}{BL}\):**
   - The Angle Bisector Theorem states that the ratio of the two segments created by the angle bisector is equal to the ratio of the other two sides of the triangle.
   - Applying the Angle Bisector Theorem to \(\triangle ABC\) with angle bisector \(\overline{CL}\):
     \[
     \frac{AB}{BL} = \frac{AC}{BC} = \frac{450}{300} = \frac{3}{2}
     \]

5. **Substitute the values back into Menelaus' Theorem:**
   - Substituting \(\frac{AB}{BL} = \frac{3}{2}\) into the Menelaus' equation:
     \[
     \frac{1}{1} \cdot \frac{3}{2} \cdot \frac{LP}{180} = 1
     \]
   - Simplifying the equation:
     \[
     \frac{3}{2} \cdot \frac{LP}{180} = 1
     \]
     \[
     \frac{3 \cdot LP}{360} = 1
     \]
     \[
     3 \cdot LP = 360
     \]
     \[
     LP = \frac{360}{3} = 120
     \]

The final answer is \(\boxed{120}\)