The correct answer is $\boxed{\text{B) is 0}}$.
The line integral $\displaystyle 2\int_P^Q \left( x \, dx + y \, dy \right)$ is the work done by a force $\mathbf{F}(x, y) = x \hat{\imath} + y \hat{\jmath}$ in moving a particle from point $P$ to point $Q$.
In this case, $P = (1, 0)$ and $Q = (0, 1)$, so the line integral is
$$2\int_P^Q \left( x \, dx + y \, dy \right) = 2\int_1^0 \left( x \, dx + y \, dy \right) = 2\int_1^0 \left( 1 \, dx + 0 \, dy \right) = 2x \bigg|_1^0 = 0.$$
The value of the line integral does not depend on the direction of the semicircle, because the force $\mathbf{F}$ is conservative. A conservative force is a force that can be written as the gradient of a potential function, $\mathbf{F} = -\nabla V$. In this case, the potential function is $V(x, y) = \frac{1}{2}x^2 + \frac{1}{2}y^2$. The work done by a conservative force in moving a particle from point $P$ to point $Q$ is equal to the change in the potential energy of the particle, which is zero in this case.