smc

Draw $MB$. Since $\angle AMB$ is inscribed on a diameter, $\angle AMB$ is $90^{\circ }$. By AA Similarity, $△APO\sim △ABM$. Setting up ratios, we get $\frac{AP}{AO}=\frac{AB}{AM}$. Cross-multiplying, we get $AP\cdot AM=AO\cdot AB$, so the answer is $\boxed{AO\cdot AB}$.