$\langle x | p \rangle = \psi_p(x)$ , What this anything to do with $\psi$ (Notation in R.Shankar) In Principal of Quantum Mechanics R.Shankar Page 137 
if we project the eigenvalue equation
$$P|p\rangle = p|p\rangle$$
onto the $X $ basis,we get
$$\langle x|P|p\rangle = p \langle  x |p \rangle$$
or,
$$-i\frac{h}{2\pi}\frac{d\psi_p(x)}{dx}=p\psi_p(x) $$
where, $$\psi_p(x)=\langle x |p \rangle.$$
They use the notation $\psi_p(x)$ for $\langle x |p \rangle$ , I know it is just a notation but $\psi$ has noting to do with  $\langle x |p \rangle$ and $p$ is labeled in $\psi(x)$(=$\langle x|\psi \rangle$ ) which is continuous.
The same is used in page 313 for $\langle x| l_z \rangle $ equating it to $\psi_{l_z}(\rho,\phi)$.
 A: You have an important mistake. 
\begin{align}
\langle x|P|p\rangle &\neq p \langle  x \rangle \\
 & = p \langle x|p\rangle \\
 & = p \frac{e^{ipx/\hbar}}{\sqrt{2\pi}}\\
 & = -i\hbar \frac{\partial}{\partial x}\frac{e^{ipx/\hbar}}{\sqrt{2\pi}}.
\end{align}
Now, for what you're missing is that $\psi(x) = \langle x|\psi\rangle$ for any wave function (state) $\psi(x)$ ($|\psi\rangle$). In this case, the state in question is $|p\rangle$ with the wave function is $\psi_p(x)$ (don't overlook the subscript!).
A: A (wave-)function $\psi$ in a Hilbert space, is in fact a vector with infinitely many components. In bra-ket notation we write it as $|\psi \rangle$. Instead the complex number $\psi(x)$ is the component of the vector in the (overcomplete) basis $|x\rangle$ which is the position basis. Such wavefunction represents a particle sitting at position $x$ with certainty. So its components themselves must be
$$
\langle x' | x \rangle = \delta(x'-x).
$$
The number $\psi(x)$, formally, is also the scalar product of $|x\rangle$ with $|\psi\rangle$. Indeed, again formally,
$$
\langle x | \psi\rangle = \int dx' \delta(x'-x) \psi(x') = \psi(x). 
$$
In its book Shankar simply uses the (common) notation $|p\rangle$ for the wavefunction $\psi_p$ since he knows it depends on the quantum number $p$.
