解: (2)$PA = PE。$理由如下：如图①，延长$EB$至点$H，$使$BH = BP，$连接$PH。$
$\because\triangle ABC$是等边三角形，$\therefore AB = BC，$$\angle ACB=\angle ABC = 60^{\circ}。$
$\because BM// AC，$$\therefore\angle ACB=\angle CBH = 60^{\circ}。$又$\because BP = BH，$$\therefore\triangle BPH$是等边三角形，$\therefore PH = BP = BH，$$\angle H = 60^{\circ}=\angle ABC=\angle APE=\angle BPH，$$\therefore\angle APB=\angle EPH，$$\therefore\triangle APB\cong\triangle EPH(ASA)，$$\therefore PA = PE。$
(3)当点$P$在$BC$上时，$BC = BP + BE；$当点$P$在线段$CB$的延长线上时，$BE = BP + BC。$理由如下：当点$P$在$BC$上时，由
(2)可知，$\triangle APB\cong\triangle EPH，$$\therefore AB = EH，$$\therefore BC = EH = EB + BH = BE + BP。$当点$P$在线段$CB$的延长线上时，如图②，在$BE$上截取$BH = BP，$连接$PH。$
$\because\triangle ABC$是等边三角形，$\therefore AB = BC，$$\angle ACB=\angle ABC = 60^{\circ}。$
$\because BM// AC，$$\therefore\angle ACB=\angle PBH = 60^{\circ}。$又$\because BP = BH，$$\therefore\triangle BPH$是等边三角形，$\therefore PH = BP = BH，$$\angle BHP = 60^{\circ}=\angle ABC=\angle APE=\angle BPH，$$\therefore\angle APB=\angle EPH，$$\angle EHP=\angle ABP = 120^{\circ}，$$\therefore\triangle APB\cong\triangle EPH(ASA)，$$\therefore EH = AB，$$\therefore BE = BH + EH = BP + BC。$
