解：$(1)∵$四边形$ABCD $的内角和为$360°，$$∠A = 90°，$$∠C=60°$
$∴∠ADC+∠ABC=360°-(∠A+∠C)= 210°$
$∵∠ADC$和$∠ABC $的平分线交于点$O$
$∴∠ADO=\frac {1}{2}∠ADC，$$∠ABO=\frac {1}{2}∠ABC$
$∴∠ADO +∠ABO=\frac {1}{2}(∠ADC+∠ABC)=105°$
∵四边形$ABOD$的内角和为$360°$
$∴∠BOD =360°-(∠ADO +∠ABO)-∠A=165°$
$(2)$设$∠ABP=x$
$∵BP $平分$∠ABC$
$∴∠PBC=∠ABP=x$
∵四边形$ABCD $的内角和为$360°，$$∠A=90°，$$∠C = 60°$
$∴∠ADC =360°- ∠A-∠C-∠ABC= 210°-2x.$
$∵∠ADC + ∠EDC = 180°$
$∴∠EDC=180°-∠ADC=2x-30°$
$∵DP $平分$∠EDC$
$∴∠PDC= \frac {1}{2}∠EDC=x-15°$
设$PB、$$CD$交于点$O$
$∵△DPO、$$△BCO $的内角和为$ 180°，$$∠DOP = ∠BOC$
$∴∠BPD + ∠PDC = ∠C + ∠PBC$
$∴∠BPD = ∠C +∠PBC -∠PDC =60°+x-(x-15°)= 75°$
$(3)DG $与$BH$不平行，理由：
连接$BD$
∵四边形$ABCD$的内角和为$ 360°，$$∠A= 90°，$$∠C= 60°$
$∴∠ADC + ∠ABC =360°-(∠A+∠C)=210°$
$∵∠CDE+∠ADC=180°，$$∠CBF+∠ABC=180°$
$∴∠CDE+∠CBF=180°×2-(∠ADC+∠ABC)=150°$
$∵DG、$$BH$分别是$∠CDE、$$∠CBF $的平分线
$∴∠GDC =\frac {1}{2}∠CDE，$$∠HBC=\frac {1}{2}∠CBF$
$∴∠GDC+∠HBC =\frac {1}{2}(∠CDE+∠CBF)=75°$
$∵△DBC$的内角和为$180°，$$∠C=60°$
$∴∠BDC +∠DBC=120°$
$∴∠GDB+ ∠HBD =120°+ 75°=195°≠180°$
$∴DG $与$BH$不平行
