What is relation between the path fibration of classifying map and the Borel construction I am trying to solve exercise 179 on Davis-Kirk:

  
*
  
*Show that given a principal $G$-bundle $E\to B$, there is a fibration
  $$E\hookrightarrow EG\times_G E\to BG$$
  where $EG\times_G E$ denotes the Borel construction. 
  
*How is this fibration related to the path fibration $$P_f\to BG$$ 
  where $f:B\to BG$ is the classifying map of the bundle $E\to B$?

The first part is trivial since $BG$ is a CW complex and we obtain a fiber bundle by Borel construction. The statement follows from the fact any fiber bundle over paracompact space is a fibration. 
However, I have no idea what exactly does the second question mean. Should we construct a morphism between these two fibrations?

P.S.
The path fibration is given by the pull-back 
$$\require{AMScd}
\begin{CD}
P_f @>>>  BG^I \\
@VVV @VVV \\
B @>{f}>>  BG
\end{CD}
$$
 A: Let $p:E\rightarrow B$ be a principal $G$-bundle and consider the Borel construction  $EG\times_G E$. As you have found there is an induced map $\pi:EG\times_GE\rightarrow BG=EG/G$ with fibre $E$ obtained by projecting to the first factor. On the other hand, you can also project to the second factor to get a map
$$q:EG\times_GE\rightarrow B=E/G.$$
Since $p$ is a locally-trivial $G$-bundle the map $q$ has the structure of a locally-trivial fibre bundle with fibre $EG$. Moreover, since $EG$ is contractible, $q$ is a (weak) homotopy equivalence, which you can check by studying its long-exact sequence of homotoyp groups.
Therefore consider the following diagram
$\require{AMScd}$
\begin{CD}
    G@>>> E @>p>> B\\
    @VV V   @VV  V@VV=V\\
    EG @>>> EG\times_GE@>q>\simeq >B\\
@VVV @VV\pi V\\
BG@>>=>BG
\end{CD}
which you can check commutes strictly. The top left-hand square is a pullback, which tells you that the fibre inclusion $E\hookrightarrow EG\times_GE$ of $\pi$ is a principal $G$-bundle. Since we know that $EG\times_GE\simeq B$, what we have achieved here is to construct an explicit classifying map for the bundle $p$, and at the same time have arranged for it to be a fibration. Namely the composite $\pi\circ q^{-1}:B\rightarrow BG$ is a classifying map for $E$, with $\pi$ a fibration. The point is that we have used only 'geometric' information to achieve this. Had we started with a map $f:B\rightarrow BG$ classifying $p$ and asked to understand its homotopy fibre $F_f$, this is exactly what we would have got.
To compare these two results directly choose a homotopy equivalence $\xi:EG\xrightarrow{\simeq}PBG$ and it together with the map $q$ to construct a map from the square
$\require{AMScd}$
\begin{CD}
    E@>>> EG \\
    @VV V   @VV  V\\
    EG\times_GE @>\pi>> BG\\
\end{CD}
to the square
$\require{AMScd}$
\begin{CD}
    F_f@>>> PBG \\
    @VV V   @VV  V\\
    B @>f>> BG\\
\end{CD}
Since both $q$ and $\xi$ are homotopy equivalences, the induced map of pullbacks $E\rightarrow F_f$ is a homotopy equivalence.
