Momentum operator in momentum representation in polar coordinates

In order to write the Schrödinger equation in $lbrace|mathbf{p}ranglerbrace$ representation you must project:

$$ ihbarfrac{d}{dt}|psi(t)rangle=H|psi(t)rangle $$

onto $|mathbf{p}rangle$.

Considering that for a particle in a scalar potential $V(mathbf{r})$ (where $mathbf{r}$ might be $(x,y,z),(r,varphi,z)$ ...) the Hamiltonian operator is given by $H=frac{1}{2m}mathbf{P}^2+V(mathbf{R})$ ($mathbf{P}$ and $mathbf{R}$ are the momentum and position operators), we have:

$$ ihbarfrac{partial}{partial{t}}langlemathbf{p}|psi(t)rangle=frac{1}{2m}langlemathbf{p}|mathbf{P}^2|psi(t)rangle+langlemathbf{p}|V(mathbf{R})|psi(t)rangle $$

At this point you have to consider that $langlemathbf{p}|psi(t)rangle=overline{psi}(mathbf{p},t)$ and also the answer to your question:

Would momentum operator be still just a mupltiplication like in 1D?

Indeed, the rule of actuation of $mathbf{P}$ operator in momentum representation is always a multiplication by the eigenvalue $mathbf{p}$. The schrodinger equation then becomes:

$$ ihbarfrac{partial}{partial{t}}overline{psi}(mathbf{p},t)=frac{mathbf{p}^2}{2m}overline{psi}(mathbf{p},t)+langlemathbf{p}|V(mathbf{R})|psi(t)rangle $$

In order to calculate the last quantity we must insert the closure relation of the momentum representation:

$$ langlemathbf{p}|V(mathbf{R})|psi(t)rangle=int{d^3}p'langlemathbf{p}|V(mathbf{R})|mathbf{p}'ranglelanglemathbf{p}'|psi(t)rangle $$

and keep in mind that for any function $F(mathbf{r})$, the matrix element of $F(mathbf{R})$ operator in momentum representation is given by:

$$ langlemathbf{p}|F(mathbf{R})|mathbf{p}'rangle=(2pihbar)^{-3/2} overline{F}(mathbf{p}-mathbf{p}') $$

The schrödinger equation in momentum representation can finally be written as:

$$boxed{ ihbarfrac{partial}{partial{t}}overline{psi}(mathbf{p},t)=frac{mathbf{p}^2}{2m}overline{psi}(mathbf{p},t)+(2pihbar)^{-3/2}int{d^3}p'overline{V}(mathbf{p}-mathbf{p}')overline{psi}(mathbf{p}',t)} $$

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$overline{V}(mathbf{p})$ is of course the Fourier transform of $V(mathbf{r})$. In the case of the potential that only depends on $r$, $overline{V}$ depends only on $p$ and could be written as:

$$ overline{V}(p)=frac{1}{sqrt{2pihbar}}frac{2}{p}int_{0}^{infty}rdrsin{frac{pr}{hbar}}V(r) $$