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$^1$ $$ G=i\hbar\frac{d}{dx} \quad\iff\quad T_a=e^{iaG/\hbar}. $$
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 eigenstates of the momentum operator are the states $\lvert p\rangle$ whose wavefunctions are$^2$ $$ \langle x\vert p\rangle=e^{ipx/\hbar} $$ 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.
$^1$this 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)
$^2$this is because in position representation the eigenvalue equation reads $$ -i\hbar\frac{d\phi_p}{dx}(x)=p\phi_p(x) $$ that has the (unique) aforementioned solution (note that they are not normalizable so they are actually generalized eigenstates)