It is not a direct answer to your question, but reading [this answer][1] (mentioned in the comments) is an excellent way to understand what is necessary for this circuit to, sort of, work as a memory.

With that understood, you can take the next step to understand that the propagation delay is important to allow the storage. It determines how long the input signal must be **strongly** pulled to high or low for the output to be able to **weakly** sustain the input. The input can only turn to *"floating"* after this delay.

Note that an ideal circuit, with zero propagation delay, would also work, given the conditions presented in the linked answer. A direct answer would then be: propagation delay is not **required** for the circuit to work as a memory but if it exists, it must be taken into account for the effective storage.

  [1]: https://electronics.stackexchange.com/a/309731/194393