A fixed horizontal cylinder has a rough surface…

Step 1. A tiny piece of rope. Take a piece that covers a small angle $d\theta$. Its ends are pulled by $T$ and $T + dT$. Each pull is tilted towards the cylinder by $d\theta/2$, so together they press the piece onto the cylinder with $dN = T\,d\theta$. On the point of slipping, friction on the piece is $\mu\,dN$, and it balances the difference of the pulls:

Step 2. Add up the pieces. $\dfrac{dT}{T} = \mu\,d\theta$. Integrate from the low-tension end to the high-tension end: $\ln\dfrac{T_1}{T_2} = \mu\theta$, so $T_1 = T_2\,e^{\mu\theta}$. If the rope is about to slip towards Q instead, friction acts the other way, and $T_2 = T_1\,e^{\mu\theta}$. Here $\mu\theta = 0.25 \times \pi = 0.785$, and $e^{0.785} = 2.193$.

Step 3. At rest. Each block hangs at rest, so the tension on its side equals its weight. The rope does not slip as long as the larger tension is at most $2.193$ times the smaller one:

Step 4. Sliding. With $m_1 = 6.0\ \text{kg}$, more than $4.39\ \text{kg}$, P goes down and Q goes up, both with acceleration $a$. The rope is light, so Step 2 still holds with sliding friction: $T_1 = 2.193\,T_2$.

Put $T_2 = m_2(g + a)$ into the first equation and solve for $a$:

Then $T_2 = 2.0\ \text{kg} \times 11.55\ \text{m/s}^2 = 23.1\ \text{N}$ and $T_1 = 2.193 \times 23.1\ \text{N} = 50.7\ \text{N}$.