$解：(1) \because P A 與 \odot O 相切于點(diǎn) A， $
$\therefore O A \perp P A . $
$\therefore \angle P A B=90^{\circ} . $
$\therefore \angle A O P=90^{\circ}-\angle P=90^{\circ}-20^{\circ}=70^{\circ} . $
$\therefore \angle B=\frac {1}{2} \angle A O C=\frac {1}{2} \times 70^{\circ}=35^{\circ} $
$(2) 連接 D B 、 O D. $
$\because 弦 A D \perp O P 于點(diǎn) E， $
$\therefore A E=D E， \widehat{A C}=\widehat{D C} . $
$\because O A=O B， A E=D E， $
$\therefore O E 為 \triangle A B D 的中位線. $
$\therefore O E=\frac {1}{2}\ \mathrm {B} D .$
$ \because O E=\frac {1}{2}\ \mathrm {C} D， $
$\therefore C D=D B . $
$\therefore \widehat{C D}=\widehat{B D}.$
$∴\widehat{AC}=\widehat{CD}=\widehat{DB}$
$\therefore \angle A O C=\angle C O D=\angle B O D=60^{\circ} .$
$ \because P A 為 \odot O 的切線，$
$\therefore \angle P A O=90^{\circ}$
$ \therefore \angle P=90^{\circ}-\angle A O P=90^{\circ}-60^{\circ}=30^{\circ}$
