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$
∴$∠BDE = ∠BPC = 90°$
∴$∠DBC = 45°$
∴$PC = PB = 7\ \mathrm {cm}$
∵$AD//BC$
∴$∠ADP = 45°$
∴$PD = PA = 3\ \mathrm {cm}$
∴$AC = BD = 7+3 = 10\ \mathrm {cm}$
∴$S_{梯形ABCD}= \frac 12AC·BD=\frac 12×10×10= 50(\mathrm {cm}^2)$