Ceva or Menelaus?

$\because FE//BC$ , $\therefore \frac{AF}{FB}=\frac{EA}{CE}$ . Apply Ceva theorem with $\triangle ABC$ and point $P$ , we get $\begin{aligned} \frac{AF}{FB}× \frac{BD}{DC}× \frac{CE}{EA}&=1 \\ BD&=DC \end{aligned}$ Apply Menelaus theorem with $\triangle ACD$ and line $BPE$ , we get $\begin{aligned} \frac{AE}{EC}× \frac{CD+DB}{BD}× \frac{DP}{PA}&=1 \\ \frac{2}{3}× \frac{2}{1}× \frac{DP}{PA}&=1 \\ 4DP&=3PA \end{aligned}$ $\begin{aligned} AD&=AP+PD \\ 4AD&=4AP+4PD \\ 28&=4AP+3AP \\ AP&=\large 4 \end{aligned}$

First, we notice that since $FE$ is parallel to $BC$ , then $\triangle AFE$ is similar to $\triangle ABC$ , and thus, $AF/FB = AE/EC = 2/3$ . Now, we use mass points.

Let $m_A = 3$

We get $m_B = m_C = 2$

$m_D = m_B + m_C = 4$

$AP/AD = m_D/(m_A + m_D) = 4/7$

$\boxed{AP = 4}$