Connect $EFGH$ as shown:

First, let's compute the area of $EFGH$. This can be done by subtracting the areas of the four corner triangles from $ABCD$. Because $$S_{\triangle{AEH}}=S_{\triangle{FCG}}=\frac{1}{2}\times 3 \times 2 = 3, S_{\triangle{EBF}}=S_{\triangle{GDF}}=\frac{1}{2}\times 1\times 4 = 2$$ We find $$S_{EFGH}=S_{ABCD}-(2\times 3 + 2\times 2)=4\times 6 - 10 = 14$$ Meanwhile, by SAS, we note that $$\triangle{AEH}\cong\triangle{CGF}\qquad\text{and}\qquad\triangle{EBF}\cong\triangle{GDH}$$ Hence $HE = GF$ and $EF=GH$. As a result, we find $EFGH$ is a parallelogram. By using the base and sum of altitudes, it is easy to conclude that the the total area of $\triangle{HPE}$ and $\triangle{FPG}$ is half of that of $EFGH$, or $$S_{\triangle{HPE}}+S_{\triangle{FPG}}=\frac{1}{2}\times 14 = 7$$ Furthermore we have $$S_{\triangle{AEF}}=S_{\triangle{FCG}}=\frac{1}{2}\times 3 \times 2 = 3\quad\text{and}\quad S_{\triangle{EPH}}=S_{AEPH}-S_{\triangle{AEF}}=5-3=2$$ It follows that $S_{\triangle{FPG}}=6-S_{\triangle{EPH}}=7-2=5$. Therefore we find $S_{PFCG}=S_{\triangle{PFG}}+S_{\triangle{FCG}}=5+3=8$.