Let's analyze propositions $p$ and $q$ to determine their truth values.

For $p$:
Suppose we take $a = 1$ and $b = -1$. We have $a > b$ and $\dfrac{1}{a} = 1 > -1 = \dfrac{1}{b}$. Therefore, there exist real numbers $a$ and $b$ satisfying the condition, so proposition $p$ is true.

For $q$:
We have $\sin x + \cos x = \sqrt{2}\sin \left(x + \dfrac{\pi}{4}\right)$. Since the maximum value of $\sin$ function is $1$, this expression is maximal when $\sin \left(x + \dfrac{\pi}{4}\right) = 1$. So, the maximum value of $\sin x + \cos x$ is $\sqrt{2}$, which is less than $\dfrac{3}{2}$. Hence, this inequality holds for all real numbers $x$, and proposition $q$ is true.

Since both propositions $p$ and $q$ are true, the compound proposition $p \land q$ is true.

Therefore, the correct answer is:
$$\boxed{A}$$