Let $Q$ be the point on $BC$ such that $\widehat{BAP}=\widehat{BPQ}$ and $\widehat{CDP}=\widehat{CPQ}$.
Since the quadrilaterals $PEAF,PFDG$ are cyclic, $$\widehat{EFG}=\widehat{PFE}+\widehat{PFG}=\widehat{PAE}+\widehat{PDG}=\widehat{BPQ}+\widehat{CPQ}=\widehat{BPC},$$
hence we just need to prove that $\frac{EF}{FG}=\frac{PB}{PC}.$ Since: $$\frac{EF}{FG}=\frac{PA\sin\widehat{EPF}}{PD\sin\widehat{FPG}}=\frac{PA\sin\widehat{BCD}}{PD\sin\widehat{ABC}}=\frac{PA\cdot BD}{PD\cdot AC}$$ we only need to prove that: $$\frac{PA\cdot PC\cdot BD}{PD\cdot PB\cdot AC}=1.\tag{1}$$ This can be seen to be equivalent to the concurrency of the lines $AB,CD,PQ$. By applying the sine theorem to the triangles $ABP$ and $CDP$ and exploiting the facts that $$AC=2R\sin\widehat{ADC}=2R\sin\widehat{ABC},\quad BD=2R\sin\widehat{BAD}=2R\sin\widehat{BCD},$$ where $R$ is the circumradius of $ABCD$, we have: $$\frac{PA\cdot PC\cdot BD}{PD\cdot PB\cdot AC}=\frac{\sin\widehat{ABP}\cdot\sin\widehat{QPC}\cdot\sin\widehat{BCD}}{\sin\widehat{PCD}\cdot\sin\widehat{BPQ}\cdot\sin\widehat{ABC}},\tag{2}$$ the Trig-Ceva condition for the concurrency of $AB,CD,PQ$ with respect to the triangle $PAB$. But this concurrency is trivial since the line through $P$ and $Q$ is the radical axis of the circumcircles of $ABP$ and $CDP$ - they are tangent in $P$ since $\widehat{BAP}=\widehat{BPQ}$ and $\widehat{CDP}=\widehat{CPQ}$.
$ABCD$ is an inscribed quadrilateral.