\(\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{BC}=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{PC}\)
\(=\dfrac{-1}{2}\overrightarrow{AB}-\dfrac{1}{2}\overrightarrow{CP}\)
=>\(2\cdot\overrightarrow{BN}=-\overrightarrow{AB}-\overrightarrow{CP}\)
=>\(\overrightarrow{AB}=-2\cdot\overrightarrow{BN}-\overrightarrow{CP}\)
\(\overrightarrow{BC}=-\overrightarrow{CB}=-2\cdot\overrightarrow{CP}\)