$A$, $B$, $C$, and $D$ are points on a circle, and segments $\overline{AC}$ and $\overline{BD}$ intersect at $P$, such that $AP=8$, $PC=1$, and $BD=6$.  Find $BP$, given that $BP<DP$.
1. Let \( BP = x \) and \( DP = 6 - x \). We are given that \( BP < DP \), so \( x < 6 - x \), which simplifies to \( x < 3 \).

2. By the Power of a Point theorem, which states that for a point \( P \) inside a circle, the products of the lengths of the segments of intersecting chords through \( P \) are equal, we have:
   \[
   AP \cdot PC = BP \cdot PD
   \]

3. Substituting the given values, we get:
   \[
   8 \cdot 1 = x \cdot (6 - x)
   \]
   Simplifying this, we obtain:
   \[
   8 = x(6 - x)
   \]
   \[
   8 = 6x - x^2
   \]
   \[
   x^2 - 6x + 8 = 0
   \]

4. Solving the quadratic equation \( x^2 - 6x + 8 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -6 \), and \( c = 8 \):
   \[
   x = \frac{6 \pm \sqrt{36 - 32}}{2}
   \]
   \[
   x = \frac{6 \pm \sqrt{4}}{2}
   \]
   \[
   x = \frac{6 \pm 2}{2}
   \]
   \[
   x = \frac{6 + 2}{2} \quad \text{or} \quad x = \frac{6 - 2}{2}
   \]
   \[
   x = 4 \quad \text{or} \quad x = 2
   \]

5. Given that \( BP < DP \), we select the solution where \( BP = 2 \) and \( DP = 4 \).

The final answer is \( \boxed{2} \).