Polyphase components Spectrum formula in Discrete time
Let's start with your equation.

$$ P_k\left( e^{j\Omega} \right) = \frac{1}{L} e^{jk\Omega/L} \sum_{p=0}^{L-1} e^{-2 \pi j k p / L} H\left( e^{j(\Omega-2\pi  p)/L} \right) $$

Do a little rearranging.

$$ P_k\left( e^{j\Omega} \right) =  e^{j\Omega\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} H\left( e^{j\Omega\frac{1}{L} }  e^{-j2\pi \frac{p}{L}} \right) \left( e^{j2\pi/L} \right)^{-kp}   \right] $$

It can be made simpler to understand by a few variable substitutions.

$$ y = e^{j\Omega} $$

$y$ is a complex value on the complex unit circle at $\Omega$ radians around the circumference.  

$$ h[p] = H\left( y^{\frac{1}{L}} e^{-j2\pi \frac{p}{L}} \right) $$

$h[]$ is a sequence of values sampled from the $H()$ function.  $y^{\frac{1}{L}}$ is a complex value on the unit circle at $1/L$th the angle of $y$.  The $e^{-j2\pi \frac{p}{L}}$ is a clockwise walk around the unit circle in $L$ even steps.  Conceptually: 

$$ e^{-j2\pi \frac{p}{L}} = \left( e^{j2\pi } \right)^{-\frac{p}{L}} = 1^{-\frac{p}{L}} $$

Therefore $h[]$ has a period of $L$.  If $p$ were to go below zero or above $L-1$, it would be a repeat pattern.

With these substitutions, your equation becomes this:

$$ P_k\left( y \right) =  y^{\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} h[p] \left( e^{j2\pi/L} \right)^{-pk} \right] $$

The part in brackets is the definition of the DFT.

$$ X[k] = \sum_{n=0}^{N-1} x[n] \left(e^{i2\pi/N}\right)^{-nk} $$

It's arranged the same way as in my answer here Fractional powers of complex numbers (DSPrelated computation)

The $\frac{1}{L}$ is a normalizing factor making the magnitudes of the DFT calculation independent of $L$.

$ y^{\frac{k}{L}} $ is a complex value $ k/L $ of the way to $y$ along the unit circle which means the results of the DFT bin are rotated that far.

So, a long answer to your question.  The purpose of the second exponential $ e^{-2\pi j k p/L} $ is to execute the DFT transform.  It is part of the definition and the reason it works the way it does.  Lots of stuff available on that if you are unfamiliar.