Why is velocity of the mass $v/ \cos θ $? Why not $2v \cos θ$? 
If we call the velocity of the mass $u$, then we can say: $u \cos θ = v$ and hence $u= \frac v{\cos θ}$. But why can't we write  $v \cos θ= u$ and add the contribution from both of the strings: $u= \frac{2v}{\cos θ}$ ? Why is it only $u= \frac v{\cos θ}$?
Please note that it is not a homework based  question. I just want someone to solve the doubt arising in my mind.
 A: Imagine what would happen if the two pulleys were very close together, so that $\theta$ is very close to zero, and $\cos\theta$ is very close to one.
In that case, the mass in the middle will rise with a speed very close to $v$ - and it rises at the same rate no matter how many of the other masses are present.  Therefore, its speed can't possibly be $\frac{2v}{\cos\theta}$.  It must be $\frac{v}{\cos\theta}$.
A: Since there are four choices, we can tell the answer immediately by letting $\theta$ go to $0$ and $\frac{\pi}{2}$. By letting $\theta$ go to $0$, we can eliminate 3) and 4), as is mentioned by @Dawood ibn Kareem; by letting $\theta$ go to $\frac{\pi}{2}$, we can eliminate 1).
Otherwise we can get this result with some calculation. In order to simplify the description, let's assume the pulleys are infinitesimal (this makes no difference to the problem). Denote the distance between $A$ and $B$ by $2d$, the distance between $A$ and $P$ by $y$, the vertical distance between $A$ and the suspension point by $x$. Now from that the string is unstretchable, we know $$\sqrt{x^2+d^2}+y=const.$$ Taking derivative with respect to time $t$, we have $$\frac{x}{\sqrt{x^2+d^2}}\dot{x}+\dot{y}=0.$$ That is $$|\dot{x}|\cos\theta=|\dot{y}|,$$ where $u=|\dot{x}|$, $v=|\dot{y}|$.
A: (a) $u \cos\theta = v$ is correct. The point of intersection of the 3 strings (which I shall call Y) is moving upwards with speed $u$. This point Y is at the end of the two strings attached to P and Q. Each string AY and BY is shortening at the speed $v$, so the component of velocity of Y along the direction of each string must be $v$. Therefore $v=u\cos\theta$ is correct.
(b) $u=v\cos\theta$ is not correct. The point at which the string is attached to pulley P (I shall call this point A) is moving with velocity $v$ at angle $\theta$ to the vertical. However, the motion of Y is not the same as the motion of A. As well as moving with speed $v$ in the direction AY, the string AY is also rotating about point A. The angle $\theta$ is changing. So the motion of Y (at the other end of the string AY) is the sum of the radial component along AY and a tangential component perpendicular to AY. Writing $u=v\cos\theta$ ignores this tangential motion of Y perpendicular to AY. 
The difference between (a) and (b) is that in (a) the point Y has no motion perpendicular to the vertical, whereas in (b) the point Y does have a component of motion perpendicular to AY.
If the section of string AY did not rotate but kept a constant angle $\theta$ with the vertical, then the points A and Y would have the same velocity, in terms of direction as well as magnitude. In this case $u=v\cos\theta$ would be correct. However, Y would not be moving vertically in this case. It would be moving along AY. It cannot move along BY and AY at the same time, because then Y would have to split apart.  
Note that if the masses P and Q do not move with the same speed $v$ (which could happen if they have different weights), or if the angles $\theta$ on each side of the vertical are different (which could happen if Y is not on the mid-line between A and B), then the velocity of Y will not be vertical. As a result, neither $v=u\cos\theta$ nor $u=v\cos\theta$ will be correct.
