Let equilateral triangle $ABC$ with side length $6$ be inscribed in a circle and let $P$ be on arc $AC$ such that $AP \cdot P C = 10$. Find the length of $BP$.
1. **Given:**
   - Equilateral triangle \(ABC\) with side length \(6\) is inscribed in a circle.
   - Point \(P\) is on arc \(AC\) such that \(AP \cdot PC = 10\).
   - We need to find the length of \(BP\).

2. **Angles in Cyclic Quadrilateral:**
   - Since \(ABCP\) is a cyclic quadrilateral, the opposite angles sum to \(180^\circ\).
   - Given that \(\triangle ABC\) is equilateral, \(\angle ACB = 60^\circ\).
   - Therefore, \(\angle APB = \angle BPC = 60^\circ\).

3. **Law of Cosines in \(\triangle APC\):**
   - Using the Law of Cosines in \(\triangle APC\):
     \[
     AC^2 = AP^2 + PC^2 - 2 \cdot AP \cdot PC \cdot \cos(60^\circ)
     \]
   - Since \(AC = 6\) and \(\cos(60^\circ) = \frac{1}{2}\):
     \[
     36 = AP^2 + PC^2 - AP \cdot PC
     \]
   - Given \(AP \cdot PC = 10\):
     \[
     36 = AP^2 + PC^2 - 10
     \]
     \[
     AP^2 + PC^2 = 46
     \]

4. **Law of Cosines in \(\triangle APB\) and \(\triangle BPC\):**
   - Using the Law of Cosines in \(\triangle APB\):
     \[
     AB^2 = AP^2 + BP^2 - 2 \cdot AP \cdot BP \cdot \cos(60^\circ)
     \]
     \[
     36 = AP^2 + BP^2 - AP \cdot BP
     \]
   - Using the Law of Cosines in \(\triangle BPC\):
     \[
     BC^2 = PC^2 + BP^2 - 2 \cdot PC \cdot BP \cdot \cos(60^\circ)
     \]
     \[
     36 = PC^2 + BP^2 - PC \cdot BP
     \]

5. **Adding the Equations:**
   - Adding the equations from \(\triangle APB\) and \(\triangle BPC\):
     \[
     36 + 36 = (AP^2 + BP^2 - AP \cdot BP) + (PC^2 + BP^2 - PC \cdot BP)
     \]
     \[
     72 = AP^2 + PC^2 + 2BP^2 - BP(AP + PC)
     \]
   - Using \(AP^2 + PC^2 = 46\):
     \[
     72 = 46 + 2BP^2 - BP(AP + PC)
     \]

6. **Using Ptolemy's Theorem:**
   - By Ptolemy's Theorem for cyclic quadrilateral \(ABCP\):
     \[
     AB \cdot PC + AC \cdot BP = AP \cdot BC
     \]
   - Since \(AB = AC = BC = 6\):
     \[
     6 \cdot PC + 6 \cdot BP = 6 \cdot AP
     \]
     \[
     PC + BP = AP
     \]
   - Therefore, \(AP + PC = BP\).

7. **Solving for \(BP\):**
   - Substituting \(AP + PC = BP\) into the equation:
     \[
     72 = 46 + 2BP^2 - BP^2
     \]
     \[
     72 = 46 + BP^2
     \]
     \[
     BP^2 = 26
     \]
     \[
     BP = \sqrt{26}
     \]

The final answer is \(\boxed{\sqrt{26}}\)