When an atom decays and its atomic number decreases by 2, we can conclude that it must have emitted α-particles which are helium atoms. If it emits a helium atom, its mass number must also decrease by 4.
Therefore, P must equal to N – 4.
We can eliminate options D, E and F.
Now, we know that atom Y decays further and its mass number P remains the same but its has a new atomic number Q. If an atom decays and its mass number remains the same but the atomic number changes, we can conclude that it must have emitted a ß-particle which are just high energy electrons. When an electron is emitted, the atomic number increase by 1.
After the first decay, R was reduced by 2 and after the second decay, R gained 1.
R – 2 + 1 = R – 1
P = N – 4
Q = R -1
This corresponds to option B.