a) Có \(\overrightarrow {MA}  = 2\overrightarrow {MB}  \Leftrightarrow  - \overrightarrow {AM}  = 2\left( {\overrightarrow {AB}  - \overrightarrow {AM} } \right) \Leftrightarrow \overrightarrow {AM}  = 2\overrightarrow {AB} \)

\(\begin{array}{l}
\overrightarrow {NA}  + \overrightarrow {NC}  = \overrightarrow 0  \Leftrightarrow  - \overrightarrow {AN}  + \overrightarrow {AC}  - \overrightarrow {AN}  = \overrightarrow 0  \Leftrightarrow \overrightarrow {AN}  = \frac{1}{2}\overrightarrow {AC} \\
2\overrightarrow {PB}  + \overrightarrow {PC}  = \overrightarrow 0  \Leftrightarrow 2\left( {\overrightarrow {AB}  - \overrightarrow {AP} } \right) + \overrightarrow {AC}  - \overrightarrow {AP}  = \overrightarrow 0  \Leftrightarrow \overrightarrow {AP}  = \frac{2}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {AC} 
\end{array}\)

b) Chứng minh M, N, P thẳng hàng

\(\begin{array}{l}
\overrightarrow {MN}  = \overrightarrow {AN}  - \overrightarrow {AM}  = \frac{1}{2}\overrightarrow {AC}  - 2\overrightarrow {AB}  =  - 2\overrightarrow {AB}  + \frac{1}{2}\overrightarrow {AC} \,\,\left( 1 \right)\\
\overrightarrow {MP}  = \overrightarrow {AP}  - \overrightarrow {AM}  = \frac{2}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {AC}  - 2\overrightarrow {AB}  =  - \frac{4}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {AC} \,\,\left( 2 \right)
\end{array}\)

Từ (1), (2) \( \Rightarrow \overrightarrow {MN}  = \frac{3}{2}\overrightarrow {MP}  \Rightarrow \overrightarrow {MN} ,\overrightarrow {MP} \) cùng phương nên M, N, P thẳng hàng.