Let the intersection point of lines $DP$ and $BC$ be $Q$. Let $PD=x$ and $PQ=y$. Thus, $QB=\dfrac{2y}{x}$. By Angle Bisector Theorem in $\triangle DQC$, we have $$\dfrac{x(x+y)}{x+2y}=3\Longleftrightarrow x(x+y)=3x+6y$$Let the midpoint of segment $BC$ be $M$. $MD^2=10$ and $$MD^2=10=DQ\cdot DC-QM\cdot MC=3(x+y)-\dfrac{2y}{x}-1\Longleftrightarrow 11x=3x(x+y)-2y$$Since $x(x+y)=3x+6y$, we conclude as $$11x=3x(x+y)-2y=9x+18y-2y=9x+16y\Longleftrightarrow \dfrac{AP}{BP}=\dfrac{x}{y}=8$$which implies $BP=1/3$.