$解：(2)过点D作DE//AC交BC的延长线于点E，如图$
$∵AD//BC(已知)，即AD//CE$
$∴四边形ACED是平行四边形$
$∴AC = DE$
$又∵四边形ABCD是等腰梯形$
$∴AC = BD$
$∴BD = DE$
$∵DE//AC，AC⊥BD$
$∴∠BDE = ∠BPC = 90°$
$∴∠DBC = 45°$
$∴PC = PB = 7\ \mathrm {cm}$
$∵AD//BC$
$∴∠ADP = 45°$
$∴PD = PA = 3\ \mathrm {cm}$
$∴AC = BD = 7+3 = 10\ \mathrm {cm}$
$∴S_{梯形ABCD}= \frac 12AC·BD=\frac 12×10×10= 50(\mathrm {cm}^2)$
