The other answer are all correct, but I want to give another, hopefully clearer, outlook. Momentum, in any context, can be *defined* as the quantity that is conserved as a consequence of translational symmetry.

In quantum mechanics the translation operator is defined by
$$
(T_a \psi)(x)\equiv\psi(x-a)
$$
and its generator is*
$$
G=i\hbar\frac{d}{dx} \quad\iff\quad T_a=e^{iaG/\hbar}.
$$

If the system is translationally symmetric the operator $G$ is conserved, as known from the general theory of quantum mechanics. We conventionally consider its opposite as the momentum operator.

The (generalized) eigenstates of the momentum operator are the states $\lvert p\rangle$ whose wavefunctions are
$$
\hat p\lvert p\rangle=p\lvert p\rangle \quad\iff\quad\langle x\vert p\rangle=e^{ipx/\hbar}
$$
and therefore probability of a generic state $\vert\psi\rangle$ to have a momentum $p$ is
$$
\langle p\vert\psi\rangle=\int\!dx\,\langle p\vert x\rangle\langle x\vert\psi\rangle =\int\!dx\,e^{-ipx/\hbar}\psi(x).
$$
This result basically says that the probability density of the momentum is the Fourier transform of the probability density of the position (loosely speaking, since the probabilities are their $|\cdot|^2$). This is how position and momentum are related through the Fourier transform.

---

(* this it follows from
\begin{align}
(e^{iaG/\hbar}\psi)(x)&=\left(e^{-a\frac{d}{dx}}\psi\right)(x) \\&= \left[\left(1-a\frac{d}{dx}+a^2\frac{d^2}{dx^2}+\dots\right)\psi\right](x) \\&= \psi(x)-\frac{d\psi}{dx}(x)\,a+\frac{d^2\psi}{dx^2}(x)\,a^2+\dots \\&= \psi(x-a)
\end{align}
which is strictly valid for analytic functions.)