Eigenvalue of a Hermitian operator are always real. A contradiction f(x) = $e^{-kx}$
$P_x$f(x) = -kih$e^{-kx}$
Hence, eigenvalue = -ikh
 A: The point is that $P= -i \frac{d}{dx}$ does not admit eigenvectors at all in any suitable subspace  of smooth functions  of the Hilbert space $L^2(\mathbb{R}, dx)$ so that, strictly speaking, the question is meaningless. 
Eigenfunctions can be however defined extending the definition of the  domain (and the image) of the operator outside the Hilbert space. If the domain and the image  are extended to $C^\infty(\mathbb{R})$, then $e^{kx}$ are well-behaved eigenfunctions for every $k\in \mathbb R$ (or also $\mathbb C$) and there is no reason to reject them.
However, we are here interested in extensions which are useful in QM. Therefore, the choice of an extension  has to be physically motivated. In QM, if one adopts the spectral decomposition machinery of selfadjoint operators in the sense of rigged Hilbert spaces (Gelfand-Vilenkin's approach), a suitable extension of the domain and image of $P$ is the Schwartz space of distributions ${\cal S}'({\mathbb R})$. With this extension, the extended operator $P$ ceases to be Hermitian since the space is no longer Hilbert.  
The rejection of the eigenfunction $e^{-kx}$, with real $k$, is here due to the fact that Schwartz distributions which are functions are necessarily polynomially bounded and $e^{kx}$ is not. So it cannot be used in Gelfand-Vilenkin's machinery. Conversely $e^{ikx}$, with $k \in \mathbb R$, can be accepted (as everybody knows).
I stress that there it is by no means necessary to use the rigged Hilbert space  approach (though it is very effective as every physicist knows!). Within standard von Neumann's approach to spectral decomposition which is completely enough to formulate QM, $P$ has no eigenvectors.
A: The problem with the state is that it is completely unphysical. One of the postulates of the Quantum Mechanics is that when decomposed into a sum of eigenvectors of an observable (e.g. $\hat{x}$), the norm of each coefficient of each eigenvector is equal to the probability of measurement outcome. For your state the probabilities associated to $\hat{x}$ eigenstates would be infinite, and cannot sum to 1.
Completely the same argument applies to $\hat{p}$, and the $\hat{p}$ is Hermitian only for those states that can be decomposed into a sum of $\hat{p}$ eigenstates ($e^{-ipx}$ for real $p$ and $x$). Clearly the wavefunction from your question cannot be decomposed this way.

Another way to see the issue is that $\hat{p}$ is defined through commutation relations with $\hat{x}$:
\begin{equation}
\left[ \hat{x}, \hat{p} \right] = i \hbar
\end{equation}
and (more importantly for your question) as a Noether current and generator of translations:
\begin{align}
|\phi\rangle & = e^{-i \frac{\hat{p}}{\hbar} x_0} |\psi\rangle, \\
\langle \phi | \hat{x} | \phi \rangle & =
  \langle \psi | \left( \hat{x} + x_0 \right) | \psi \rangle.
\end{align}
However your state cannot be translated, for example try computing an expectation value
$\langle x \rangle$ for original and translated states. Thus the momentum operator does not have to exist or even take the form of derivative.

A: The function $e^{-kx}$ is not an eigenfunction of $P_x$. Wave functions can behave like this only in a bounded region of space. This boundary breaks the translation symmetry, which is required for momentum conservation as stated by Noether's theorem. Only the sum of the momentum of the particle described by $e^{-x} $ and the system that embodies the boundary, for example a potential well, is conserved. 
