In the diagram, PQ || RS, QPT = 70°, PTR = 30° and TRS = x°. Find the value of x.

Let this line be $L_T$. Let $A$ be a point on $L_T$ to the left of $T$, and $B$ be a point on $L_T$ to the right of $T$. So, $A-T-B$ is a straight line parallel to $PQ$ and $RS$.

Since $PQ \parallel L_T$, the alternate interior angles $\angle QPT$ and $\angle PTA$ are equal.

Since $L_T \parallel RS$, the alternate interior angles $\angle RTB$ and $\angle TRS$ are equal.

From the diagram, the angle $\angle PTR$ is the sum of the angles $\angle PTA$ and $\angle RTB$. This is because $PT$ and $TR$ are on opposite sides of the auxiliary line $L_T$ relative to the angles formed with $L_T$.

Substitute the known values:

This result is negative, which indicates that the initial assumption about the relationship between the angles at $T$ was incorrect based on the visual representation of the angles.

Let's re-evaluate the relationship of the angles at $T$.
The angle $\angle PTA$ is formed by $PT$ and the line $L_T$ (left side).
The angle $\angle RTB$ is formed by $RT$ and the line $L_T$ (right side).
The angle $\angle PTR$ is formed by $PT$ and $TR$.

Looking at the diagram, the angle $\angle PTR$ is between $PT$ and $TR$.
The angle $\angle PTA$ is formed by $PT$ and the line $L_T$.
The angle $\angle RTB$ is formed by $TR$ and the line $L_T$.

It appears that the angle $\angle PTA$ is composed of $\angle PTR$ and $\angle RTB$.
This means that $PT$ and $TR$ are on the same side of the line $L_T$.
Let's check the diagram again. $P$ is above $T$. $R$ is below $T$.
The line $PQ$ is above $L_T$. The line $RS$ is below $L_T$.
So $PT$ goes from $P$ (above $L_T$) to $T$ (on $L_T$).
$TR$ goes from $T$ (on $L_T$) to $R$ (below $L_T$).
This means $PT$ and $TR$ are on opposite sides of $L_T$.

Let's use the method of extending $PT$ to intersect $RS$.

Since $PQ \parallel RS$, and $PK$ is a transversal, the alternate interior angles $\angle QPT$ and $\angle TKR$ are equal.

The angles inside $\triangle TRK$ are $\angle KTR$, $\angle TKR$, and $\angle TRK$.
We are given $\angle PTR = 30^\circ$. This is the same as $\angle KTR$.
So, $\angle KTR = 30^\circ$.
We found $\angle TKR = 70^\circ$.

The sum of angles in a triangle is $180^\circ$.

The angle $\angle TRS = x^\circ$ is the angle formed by the line segment $TR$ and the line $RS$.
Since $K$ lies on the line $RS$, the angle $\angle TRK$ is the same as $\angle TRS$.
Therefore, $x = 80^\circ$.

Comparing this result with the given options:
A) 30
B) 80
C) 70
D) 100

The calculated value $x=80^\circ$ matches option B.

The final answer is $\boxed{\text{B}}$.