One. (40 points) Given $0<\alpha<\beta \leqslant \frac{\pi}{2}$. On the unit circle $\odot O$, there are two fixed points $A$ and $B$, with $\angle A O B=2 \alpha$. $M$ is a moving point on $\odot O$. On the line $M A$, there exists a point $P$ such that $\angle A P K=180^{\circ}-\beta$ (where $K$ is the midpoint of side $M B$). Try to find the maximum value of $A P$.
As shown in Figure 3, on the major arc $\overparen{A B}$, take point $C$ such that $\angle 4 O C=2 \beta$. Take the midpoint $D$ of $B C$, connect $A D$ and extend it to intersect $\odot O$ at point $E$. Draw $C F / / D A$ to intersect $\odot O$ at another point $F$.
By the midline theorem, we know $C M / / D K$.
Below, we discuss point $M$ in different cases.
(1) When $M=M_{1} \in \overparen{C E}$, let $K_{1} D$ intersect $M_{1} A$ at point $P_{1}$ (where $K_{1}$ is the midpoint of $B M_{1}$).
By $\angle C M_{1} A=\frac{1}{2} \angle A O C=\beta$
$\Rightarrow \angle M_{1} P_{1} K_{1}=\beta \Rightarrow \angle A P_{1} K_{1}=180^{\circ}-\beta$.
Since $\beta \in\left(0, \frac{\pi}{2}\right]$, we have $P_{1}=P$.
Therefore, $\angle A P D=180^{\circ}-\beta$.
(2) When $M=M_{2} \in \overparen{C F}$, similarly, there exists a point $P$ on $M A$ such that
$$
\angle A P K=\angle A P D=180^{\circ}-\beta.
$$
(3) When $M=M_{3} \in \overparen{A F}$, let $K_{3} D$ intersect $M_{3} A$ at point $P_{3}$.
$$
\begin{array}{l}
\text { By } \angle F A E=\angle C E A=\frac{1}{2} \angle A O C=\beta \text {, we know } \\
\angle M_{3} A D=\angle M_{3} A E>\beta. \\
\text { Also, } \angle M_{3} P_{3} D=180^{\circ}-\angle A M_{3} C=\beta<\angle M_{3} A D,
\end{array}
$$
then point $P_{3}$ lies on the extension of $M_{3} A$, and
$$
\angle A P_{3} K_{3}=\angle A P D=\beta.
$$

Thus, the point $P$ that satisfies $\angle A P K_{3}=180^{\circ}-\beta$ lies on the circumcircle of $\triangle A P_{3} K_{3}$. However, the two intersection points of $M_{3} A$ with this circle are $A$ and $P_{3}$, so there is no point $P$ on the line $M_{3} A$ that satisfies
$$
\angle A P K_{3}=180^{\circ}-\beta.
$$
(4) When $M=M_{4} \in \overparen{A B}$, similarly, there is no point $P$ that satisfies the condition.
(5) When $M=M_{5} \in \overparen{B E}$, similarly, there exists a point $P$ on $M A$ such that
$$
\begin{array}{l}
\angle A P K=180^{\circ}-\angle A P D=180^{\circ}-\beta \\
\Rightarrow \angle A P D=\beta.
\end{array}
$$

In summary, when $M \in$
$\overparen{B E C}$, there exists a point $P$ that satisfies the condition.
As shown in Figure 4, draw $C M_{0}$
$\perp A E$ to intersect $\odot O$ at point $M_{0}$, and $D P_{0} / / C M_{0}$ to intersect
$A M_{0}$ at point $P_{0}$. Then $\angle A P_{0} D=\beta$.
From the above proof, the point $P$ that satisfies $\angle A P K=180^{\circ}-\beta$ lies on the circle with diameter $A P_{0}$. Therefore,
$$
\begin{array}{l}
A P_{\max }=A P_{0}=\frac{A D}{\sin \beta} \\
=\frac{1}{2 \sin \beta} \sqrt{2\left(A B^{2}+A C^{2}\right)-B C^{2}} \\
=\frac{1}{\sin \beta} \sqrt{2\left(\sin ^{2} \alpha+\sin ^{2} \beta\right)-\sin ^{2}(\alpha+\beta)}.
\end{array}
$$