If the tangent to the curve $$y = x + \sin y$$ at a point $$(a, b)$$ is parallel to the line joining $$(0, \frac{3}{2})$$ and $$(\frac{1}{2}, 2)$$, then:

We are given the curve $$y = x + \sin y$$ and consider a point $$(a,b)$$ on it. To find the slope of the tangent, we differentiate implicitly with respect to $$x$$.

First write the equation again so that the differentiation step is clear:

$$y = x + \sin y.$$

Differentiating both sides with respect to $$x$$, we remember that $$\dfrac{d}{dx}(y)=\dfrac{dy}{dx}$$ and that $$\dfrac{d}{dx}(\sin y)=\cos y \dfrac{dy}{dx}$$ because $$y$$ itself depends on $$x$$. Hence

$$\frac{dy}{dx} = 1 + \cos y \,\frac{dy}{dx}.$$

Now we collect the $$\dfrac{dy}{dx}$$ terms on the left:

$$\frac{dy}{dx} - \cos y \,\frac{dy}{dx} = 1.$$

Factor out $$\dfrac{dy}{dx}:$$

$$\left(1 - \cos y\right)\frac{dy}{dx} = 1.$$

So, solving for the derivative, we obtain

$$\frac{dy}{dx} = \frac{1}{1 - \cos y}.$$

At the specific point $$(a,b)$$ on the curve, the slope of the tangent (call it $$m_T$$) is therefore

$$m_T = \frac{1}{1 - \cos b}.$$

The problem states that this tangent is parallel to the line joining the two points $$\bigl(0,\tfrac32\bigr)$$ and $$\bigl(\tfrac12,2\bigr).$$ We first compute the slope of that line. The two-point slope formula is

$$m = \frac{y_2 - y_1}{x_2 - x_1}.$$

Substituting $$\bigl(x_1,y_1\bigr)=(0,\tfrac32)$$ and $$\bigl(x_2,y_2\bigr)=\bigl(\tfrac12,2\bigr)$$, we get

$$m = \frac{2 - \tfrac32}{\tfrac12 - 0} = \frac{\tfrac12}{\tfrac12} = 1.$$

Because the two lines are parallel, their slopes are equal. Thus

$$m_T = 1.$$

So we set

$$\frac{1}{1 - \cos b} = 1.$$

Cross-multiplying gives

$$1 = 1 - \cos b,$$ $$\cos b = 0.$$

The solutions of $$\cos b = 0$$ are

$$b = \frac{\pi}{2} + k\pi,\quad k \in \mathbb{Z}.$$

Next, we must relate $$a$$ and $$b$$ using the original curve equation $$y = x + \sin y.$$ At the point $$(a,b)$$ we substitute:

$$b = a + \sin b.$$

Rearrange to express $$a$$ in terms of $$b$$:

$$a = b - \sin b.$$

Because $$b = \dfrac{\pi}{2} + k\pi,$$ we note that $$\sin b = \sin\!\Bigl(\dfrac{\pi}{2}+k\pi\Bigr) = (-1)^k \cdot 1.$$ Hence $$\sin b = \pm 1,$$ and therefore

$$a = b \mp 1.$$

Taking the absolute difference, we have

$$|b - a| = |\,\sin b\,| = 1.$$

This precisely matches Option B. No other listed option is guaranteed for every integer $$k$$.

Hence, the correct answer is Option B.