You want to know what $$\hat p\psi (x)$$ is. In
In matrix algebra it's $$\left\langle x \right|\hat p\left| \psi \right\rangle $$ in x representation. In
$$\left\langle x \right|\hat p\left| \psi \right\rangle $$
In wave mechanics it's $$\hat p\psi (x) = {\hbar \over i}{\partial \over {\partial x}}\psi (x) = {\hbar \over i}{\partial \over {\partial x}}\left\langle {x}
\mathrel{\left | {\vphantom {x \psi }}
\right.}
{\psi } \right\rangle $$ . They
$$
\hat p\psi (x) = {\hbar \over i}{\partial \over {\partial x}}\psi (x) = {\hbar \over i}{\partial \over {\partial x}}\left\langle {x}
\mathrel{\left | {\vphantom {x \psi }}
\right.}
{\psi } \right\rangle \, .
$$
They should be equal so that $$\langle x|\hat p\left| \psi \right\rangle = {\hbar \over i}{\partial \over {\partial x}}\left\langle {x}
\mathrel{\left | {\vphantom {x \psi }}
\right.}
{\psi } \right\rangle $$.
Although$$
\langle x|\hat p\left| \psi \right\rangle = {\hbar \over i}{\partial \over {\partial x}}\left\langle {x}
\mathrel{\left | {\vphantom {x \psi }}
\right.}
{\psi } \right\rangle \, .
$$
\begin{align} P \psi(x) &= \langle x | P | \psi \rangle \\ &= \int \langle x | P x' \rangle \langle x' | \psi \rangle \, dx \\ &= i \hbar \int \delta'(x' - x) \psi(x') \, dx' \\ &= -i \hbar \left.\frac{\partial \psi(x')}{\partial x'} \right \rvert_{x'=x} \\ &= -i \hbar \frac{\partial \psi(x)}{\partial x} \end{align}
Although I always forget the derivation in the picturequote box above I remember they should be euqalequal.
