解：连接AC．∵AB为直径，∴∠ACB=90°．∵∠DPB=∠APC=60°，∴∠CAP=30°．∵D是弧BC的中点，∴∠CAP=∠BAD，∠CAB=60°．∴∠ABC=30°=∠ADC．则tan∠ADC= 3 3 ．故选C．
