It depends on your definition of bounded state. If a bounded state is just a smooth, rapidly vanishing for $|x|\to \infty$, wavefunction, the the claim is generally false.

Indeed, consider a smooth rapidly vanishing wavefunction $\psi$, mathematically a function of the Schwartz space ${\cal S}(\mathbb{R}^n)$. The Fourier transform send this function to a function of the same space $\hat{\psi}$.  $$\langle \psi | P_j \psi\rangle = \int_{\mathbb{R}^n} \overline{\hat{\psi}(p)}p_j \hat{\psi}(p) d^np= \int_{\mathbb{R}^n} |\hat{\psi}(p)|^2 p_j d^np$$ It is easy to construct Schwartz functions such that the integral above does not vanish: If it vanishes for a certain function, by changing the function in a small neighborhood of a point far from the origin, with a localized bump, the integral ceases to vanish.

Also considering the case $n=1$ a particle on the line the integral generally does not vanish, unless for instance the function is symmetric (up to a phase) under $p\to -p$. The eigenstates of the 1D harmonic oscillator are a special example.

If the wavfunction is a normalized eigenvector of the Hamiltonian operator with special symmetries, then the thesis is true as it happens for the harmonic oscillator. However, just considering superpositions of these eigenvectors (preserving the fact that the final wavefunction rapidly vanishes) the thesis becomes false.

It depends on your definition of bounded state. If a bounded state is just a smooth, rapidly vanishing for $|x|\to \infty$, wavefunction, the the claim is generally false.

Indeed, consider a smooth rapidly vanishing wavefunction $\psi$, mathematically a function of the Schwartz space ${\cal S}(\mathbb{R}^n)$. The Fourier transform send this function to a function of the same space $\hat{\psi}$.

The Fourier transform (I assume $\hbar=1$) however moves the theory form the position picture to the momentum picture, where the momentum operator acts as a multiplication. Here things become evident:

$$\langle \psi | P_j \psi\rangle = \int_{\mathbb{R}^n} \overline{\hat{\psi}(p)}p_j \hat{\psi}(p) d^np= \int_{\mathbb{R}^n} |\hat{\psi}(p)|^2 p_j d^np$$ It is easy to construct Schwartz functions such that the integral above does not vanish: If it vanishes for a certain function, by changing the function in a small neighborhood of a point far from the origin (thus remaining in the Schwartz space) the integral ceases to vanish.

Also considering the case $n=1$ a particle on the line the integral generally does not vanish, unless for instance the function is symmetric (up to a phase) under $p\to -p$. The eigenstates of the 1D harmonic oscillator are a special example.

If the wavfunction is a normalized eigenvector of the Hamiltonian operator with special symmetries, then the thesis is true as it happens for the harmonic oscillator. However, just considering superpositions of these eigenvectors (preserving the fact that the final wavefunction rapidly vanishes) the thesis becomes false.