Why does the voltage drop at each resistor remain the same in a parallel circuit?

Voltage is defined as Joules/Coulomb. Thus the energy that each electrons carries is higher the higher the voltage translating in higher kinetic energy or higher current. If an electron moves through a resistor with high resistance it should lose more kinetic energy by bumping into other atoms giving up thermal energy. However in a parallel circuit the voltage drop remains always the same at every resistor no matter what the resistance is. I just cannot figure out why.

A voltage drop is measured between two points. Devices in parallel are connected between the same two points.

The voltage tells us the change in energy per charge. The branches which have higher resistance have fewer charges per time (lower current) flowing through them, so the electron flow is slower in larger resistance of a parallel circuit. But that doesn't really answer the question; just gives some sensibility.

The reason the voltage must be the same across parallel resistances (or any parallel circuit devices) is that the connected ends of the devices are at common electrical potentials, and voltage is simply the difference in these potentials.

If you have two resistors, $$A$$ and $$B$$, and they are arranged horizontally. In parallel, the left ends are connected to each other and connected to the positive terminal of a 9 V battery. The right ends are connected to each other and connected to the negative terminal of the same 9 V battery and also connected to an earth ground at an absolute potential of $$2000$$ V (just an example). The left ends of the resistors are both at a potential of $$2009$$ V and the right ends are both at a potential of 2000 V. The voltage across A must be $$2009-2000= 9$$ V, and the voltage across B must be $$2009-2000=9$$ V. Change the numbers however you wish, but the voltage across A will be equal to the voltage across B.

Voltage is not the same as energy. It is energy change per charge.

The fact that voltage is energy per charge does not necessarily mean that this is kinetic energy, but rather it is potential energy. If electrons would experience no friction (as opposed to your statement that they "lose more kinetic energy by bumping into other atoms"), the distinction between potential and kinetic energy becomes irrelevant because one is just being converted into the other - electrons just get faster in the electric field. This is approximately the case, for example, for an electron beam in a cathode ray tube. Consequently the kinetic energy of electrons at the end of a cathode ray tube is usually expressed in "electron volts".

However, if the electrons experience friction (aka. resistance, "lose more kinetic energy by bumping into other atoms"), their coordinated movement is continuously converted into uncoordinated movement, i.e. statistical thermal movement. This means that instead of continuously converting potential energy into macroscopic kinetic energy (which is equivalent to saying that the electrons get accelerated more and more), they end up travelling at constant speed (whereby any gained energy from the potential is almost immediately converted into heat instead of kinetic energy). Constant velocity can also be combined with electron density to express a current density. This "drag" of the electrons cannot be deduced from the voltage (aka. potential energy) and Newton's laws of mechanics alone, but it is a consequence of Ohm's law (which describes the relationship between voltage and the equilibrium velocity/current of the electrons, or in other words, energy dissipation/friction/thermal disorder while "the electrons bump into atoms"). Ohm's law is somehow comparable to viscosity in fluid mechanics, which also violates (mechanical) energy conservation.

To make a long story short: instead of losing "more kinetic energy" in a high-valued resistor, the electrons dissipate the same (potential) energy, but in the high-valued resistor (where there are more collisions with atoms) their equilibrium velocity/current is correspondingly smaller, so that they lose a smaller amount of kinetic energy on every impact (see https://en.wikipedia.org/wiki/Mean_free_path for more explanation).

As a rough analogy think about you in a city traffic jam. The voltage corresponds to your time pressure (to get to an appointment for example). If you choose a path across the city center, your ability to takeover and accelerate is limited more by the increased traffic density there. Hence, your average velocity near the city center will be slower, because you will have to stop more often behind another car. The amount of kinetic energy you lose, will be less during each individual stop. If you take a path through the outskirts, there might be more possibilities to get to your target faster, so your average velocity is higher. But if you happen to be stopped by a car before you, you will lose more kinetic energy. In any case, the pressure (which is your desire to get to the target in time) stays the same. It's not a perfect analogy, but I hope it explains the crucial points.