You need to think of the man and the chair separately. This is because if you combine the man and the person into one thing, you get a single equation with Newton's second law with two unknown values (the acceleration and the upwards tension forces).
The man has a tension force acting upwards on him, a normal force upwards on him, and his weight downwards. The chair has tension acting upwards, the same magnitude of a normal force downwards, and it's weight downwards. We can exploit that each of the tension forces are the same (which is something that is incorrect in your diagram)$^*$, as well as the fact that the man and chair will have the same acceleration. Therefore, by Newton's second law,
$$ma=T+N-mg$$ $$Ma=T-N-Mg$$
Where $m$ and $M$ are the masses of the person and the chair respectively.
Since you are given what $m$, $M$, and $N$ are, these are two equations with two unknown values. Note that the 100 lbs corresponds to $N$, not $T$. I will leave the math to you to find what $a$ and $T$ are.
Also, note that the sum of these equations is what we get by treating evening as one system that I was discussing at the beginning of this answer. $$(m+M)a=2T-(m+M)g$$
Notice how we cannot get actual values for $a$ or $T$ with this equation and the given information. We need the previous two equations to solve the problem.
$^*$ If we assume a massless rope, and a massless, frictionless pulley, then the tension throughout the string is uniform. This is why the tension forces acting on the man and the chair are the same.