quantum mechanics - Derivative of a Position Eigenket

I was flicking through Zettili's book on quantum mechanics and came across a 'derivation' of the momentum operator in the position representation on page 126. The author derived that $\langle\vec{r}|\hat{\vec{P}}|\psi\rangle = -i\hbar\vec{\nabla}\langle\vec{r}|\psi\rangle$ (I've omitted the full derivation). However, from this relationship he concluded that $\hat{\vec{P}} = -i\hbar\vec{\nabla}$. I'm sure this is very basic but why can you immediately conclude this? Surely this assumes that $\langle\vec{r}|(-i\hbar\vec{\nabla})|\psi\rangle = -i\hbar\vec{\nabla}\langle\vec{r}|\psi\rangle$. I'm not sure why this is necessarily true.

Answer

$-i\hbar\vec{\nabla}|\psi\rangle$ is not a valid notation. The nabla operator is defined in three-dimensional Euclidean space, not the in the Hilbert space of quantum states.

When the author says $\hat{\boldsymbol{P}}=-i\hbar\vec{\nabla}$ he does not mean the momentum operator defined in the state space, but the space of wavefunctions. Then $\hat{\boldsymbol{P}}\psi(\boldsymbol{r})=-i\hbar\vec{\nabla}\psi(\boldsymbol{r})$.