Solution:
Let the tangent line to $k$ at $D$ meet the ray $\overrightarrow{O B}$ at point $S$. Then the lines $S D$, $B P$ and $O A$ are parallel and therefore $\triangle O B M \sim \triangle O S D$ and $\triangle D P M \sim \triangle D A O$. It follows that $\frac{B M}{S D} = \frac{O B}{O S}$ and $\frac{M P}{O A} = \frac{D P}{D A}$, i.e. $B M = \frac{O B \cdot S D}{O S}$ and $M P = \frac{O A \cdot D P}{D A}$.

Since $O A = O B$ and $\frac{B S}{O S} = \frac{D P}{D A}$ (we use the equality $S D = S B$), we obtain $B M = M P$, implying that the desired ratio equals $1 : 2$.