How do nuclear charge and shielding effect the periodic trend and group trend of ionization energy?

You need to understand one thing about shielding of the nuclear charge: whenever you start a new period of the periodic table you are starting over! The inert gas shell that you are adding to effectively cancels out the corresponding nuclear charge for the protons that make up the inert gas element. Add an additional proton on top of the inert gas configuration, and the corresponding s electron thinks it is "seeing" only a +1 nuclear charge.

For helium, the two electrons in the 1s shell completely cancel out the +2 nuclear charge. For neon the ten electrons in the 1s, 2s, and 2p shells completely cancel out the +10 nuclear charge of the 10 protons in the nucleus.

For lithium, the single 2s electron is "seeing" or "attracted to" a +3 charge but two of those three positive charges have been cancelled out by the two 1s electrons. So the lithium 2s electron only "sees" one positive charge!

Same thing happens with sodium. The single 3s electron only "sees" a +1 charge even though there are 11 protons at the nucleus because ten of those protons are shielded by the electrons in the closed valence shell of neon.

So how does this get us to ionization energy? With beryllium or magnesium the 2 s-electrons both "see" an effective nuclear charge of +2, and are thus more attracted to (or harder to remove/ionize from) the nucleus.. Similarly, the boron and aluminum valence electrons "see" an effective nuclear charge of +3 and are even harder to remove!

There are two additional things to finish this explanation.

Distance: the farther the electron is from the nucleus, the weaker its attraction, or the easier it is to remove/ionize. As you add inert gas shells, you are putting the next set of electrons farther and farther away from the nucleus. So as you go down the Group, you will see ionization potentials decrease.

Finally, We talked about boron and aluminum. But those were adding p-electrons on top of the filled s-orbital. It turns out that the filled s-orbital will slightly shield the nuclear charge. It is enough that there will be a noticeable decrease in ionization potential going from beryllium to boron and from magnesium to aluminum. You can see the same thing when you half-fill the p orbitals.