Solution:

Note that $EF$ is a midline of triangle $APB$, so $EF$ is parallel to $AB$ and $EF = \frac{1}{2} AB = 14 = CD$. We also have that $EF$ is parallel to $CD$, and so $CDEF$ is a parallelogram. From this, we have $EP = PC$ as well, so $\frac{CE}{CA} = \frac{2}{3}$. It follows that the height from $C$ to $EF$ is $\frac{2}{3}$ of the height from $C$ to $AB$. We can calculate that the height from $C$ to $AB$ is $12$, so the height from $C$ to $EF$ is $8$. Therefore $CDEF$ is a parallelogram with base $14$ and height $8$, and its area is $14 \cdot 8 = 112$.