The thing of it all is this:

There are two kinds of eigenfunctions of hermitian operators. The ones which admit discrete spectra (eigenvalues are separated from one another) and the other, continuous spectra (eigenvalues fill out an entire range). If the spectra is continuous, then they **DO NOT** represent possible wavefunctions (only a linear combination of them.....yes a gaussian wavepacket kind of thing may be normalizable). In case of momentum operators, $$\frac{\hbar}{i}\frac{\partial f_p(x)}{\partial x} = pf_p(x)$$....$f_p(x)$ is some momentum eigenfunction....
solving the above gives $$f_p(x) = Ae^{\frac{ipx}{\hbar}}$$
which is not a square integrable one.

But since momentum is an observable, we only take real values of p and use dirac orthonormality by $$\int_{-\infty}^{\infty}f^*_{p^\prime}(x)f_p(p)dx = |A^2|\int_{-\infty}^{\infty}e^\frac{(p^\prime-p)x}{\hbar}dx = |A^2|2\pi\hbar\delta(p-p^\prime)$$, and then picking $A=\frac{1}{\sqrt{2\pi\hbar}}$, we have $$\langle f_{p^\prime}|f_p\rangle = \delta(p-p^\prime)$$......

Now, this means that the eigen functions of momentum are sinusoidal (this itself is unrealizable since any TRUE or PERFECT sine wave has to extend from $-\infty$ to $\infty$)

But there is **no** such thing as **a particle with definite momentum**, courtesy heisenberg uncertainty principle........also implies that measurement cannot collapse a wavefunction to an eigenstate with a perfectly defined momentum

This is why we make a normalizable wavepacket with a narrow range of momenta.....to make the whole thing physically realizable. None of the eigenfunctions of $\hat p$ live in hilbert space but those with real eigenvalues (wavepackets) and which are dirac normalizable do. They (eigenfuntions of $\hat p$) do not represent possible physical states but are very useful in problems like scattering from a potential hill or a barrier.

Reference :- Griffiths, Introduction to Quantum Mechanics