These problems relate to constraints. This particular subset of costraint problems will fall under Atwood machines(Pulley problems), under Laws of Motion, under Mechanics.
These aren't usually standalone problems per se, but constraints are required for one to be able to solve pulley problems.
Basically, these do with the 'conservation of length of string'. Lets take a simple two-block-one-pulley system. The movable parts in it are the two blocks. Now, lets assume each one moves a distance $s_1,s_2$. Assume that by some mystical method, only one block moved. Write down the (signed) elongation in the string required. Do the same for the other block. Now, for the block which moves up, the elongation will be $-s_1$ (negative since its a contraction). For the block moving down, it will be $+s_2$. Since the string isn't stretchy, the net elongation should be zero. So $-s_1+s_2=0\implies s_1=s_2$. Ok, we sort of knew that. But this method is scalable, unlike the intuitive method of staring at it for a while. We can differentiate this to get relations between velocities, accelerations,jerks,jounces, whatever.
Note that in the given problem, your pulley moves as well, and it elongates the string thrice as much as its motion. You can see the constraint relation peeking out in the nearby equation (dunno what the $c$ and $T$ are)
There's a shortcut method to this, which is even more scalable.
Assiume massless pulleys(this method works for massive pulleys as well, as it is a geometric relation in the end)
For each string in the system, give it a tension $T_i$. Remember, tension in a string is constant, even if it wrapped around a (masless) pulley.
Note that there is a string between the lowermost pulley and the block B as well, even if it is not shown in the diagram. All connections should be done via a string(except connections with walls, they can be whatever)
Get relations between the tensions. Remember that net force on a masless pulley must be zero, and tension always acts away from the object.
Now, since net work done by internal forces must be zero, and tension is an internal force, then at every edge of the string (where a pulley is not connected), we can calculate work by simply multiplying tension with the displacement(giving it a negative sign if they are opposite in direction); adding the works, and equating to zero.
In your problem, if $T$ is the tension of the large string and $T_1$ is the tension in the small(nonexistant) string, then by equating forces about the lower pulley, we get $T+T+T=T_1$. Taking the same signs and symbols for displacement in the diagram, and applying virtual work, we get $T_1S_b+TS_c=0$. One can see that this simplifies to the constraint relation.
This looks long, but thats because I explained the steps. With practice, all you have to do is choose a string, give it tension $T$, and write all other tensions in terms of that tension. Then you mark the displacements and can "read off" the constraint relation no matter how complicated it is.