The Projection Postulate states that if we have an observable $O$ with discrete spectrum $\{\lambda_i\}$, after a measure in a system resulting in the eigenvalue $\lambda_a$, the initial system $|\psi_i\rangle$ is reducted to the state
$$|\psi_f\rangle=\frac{P_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}$$
where $P_a$ is the projection on the eigenspace of the eigenvalue $\lambda_a$. Any repeated measurement of $O$ in this state without time evolution will yield the same result $\lambda_a$, since
$$O|\psi_f\rangle=\frac{OP_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}=\frac{\lambda_aP_a|\psi_i\rangle}{\|P_a|\psi_i\rangle\|}=\lambda_a|\psi_f\rangle$$
i.e., $|\psi_f\rangle$ is an eigenvector of $O$ with eigenvalue $\lambda_a$.
In the case of continuous spectrum, there is some complications, since the eigenvectors are not normalizable and hence are not acceptable states. But as the measurement of an exact value of position, for example, cannot be made with arbitrary precision, what is really being measured is the projection over some region of the spectrum, like the dimension of the detector. In that case, say the particle is measured in the region $[a,b]$. By the projection postulate, the final state will be (in position representation, ignoring the normalization):
$$\psi_f(x)=P_{[a,b]}\psi_i(x)=\chi_{[a,b]}\psi_i(x)$$
where $\chi_{[a,b]}$ is the characteristic function of $[a,b]$. That is, the final wavefunction will be the restriction of the initial one to the interval $[a,b]$. Realize that we still have an uncertainty in the momentum as expected, but if the system does not evolve with time, any other measurement of position, will still be restricted to $[a,b]$, since the final wavefunction support is in that region.
The average vector momentum of an electron bound to an atom is exactly zero. (Otherwise, the electron would leave the atom!)
The average magnitude of the momentum can't be zero, because of the uncertainty principle. So Feynman is using the approximation $\vec p = \vec 0 + \Delta p \hat p$, where the magnitude $\Delta p$ comes from the uncertainty principle and the unit vector $\hat p$ points in a completely random direction.
As for your second question, you're almost there. The kinetic energy does become larger for an electron nearer the nucleus — and, thanks to the uncertainty principle, so does the momentum! It has to be this way because the kinetic energy is approximately $T=p^2/2m.$
Best Answer
Motion does not cease at absolute zero if the system you are looking at has a zero point energy.
In many systems, e.g. crystals, at low temperatures the atoms/molecules behave as harmonic oscillators, and the energy of a harmonic oscillator cannot be reduced to zero: there is always some minimum energy called the zero point energy. This means that at absolute zero the atoms in a crystal will not be stationary. There will be a small vibration corresponding to the zero point energy. This is most obvious for light atoms like Helium where the zero point energy is enough to keep the system liquid, so even at absolute zero Helium will not solidify unless it's put under pressure.
The situation is different for a free particle. In that case, at absolute zero the momentum is zero but then we have no knowledge about where the particle is (i.e. $\Delta x = \infty$). If we want to measure where the particle is we have to put some energy in, but then of course the system is no longer at absolute zero and the momentum is now non-zero.