1. Represent the two A.P. sets

Since each set has three numbers in A.P. and the sum is $36$, the middle term must be $12$.

So we can write:

• Set $A: (12-d),\ 12,\ (12+d)$
• Set $B: (12-D),\ 12,\ (12+D)$

Given: $D=d+3,\quad d>0$

2. Find the products $p$ and $q$

For set $A$: $p=(12-d)\cdot 12 \cdot (12+d)=12(144-d^2)$

So, $p=1728-12d^2$

For set $B$: $q=(12-D)\cdot 12 \cdot (12+D)=12(144-D^2)$

So, $q=1728-12D^2$

3. Use the condition $\dfrac{p+q}{p-q}=\dfrac{19}{5}$

First compute: $p+q=(1728-12d^2)+(1728-12D^2)=3456-12(d^2+D^2)$

$p-q=(1728-12d^2)-(1728-12D^2)=12(D^2-d^2)$

Since $D=d+3$, $D^2-d^2=(d+3)^2-d^2=6d+9$

Hence, $p-q=12(6d+9)=72d+108=36(2d+3)$

Now, $d^2+D^2=d^2+(d+3)^2=2d^2+6d+9$

So, $p+q=3456-12(2d^2+6d+9)$ $=3456-24d^2-72d-108$ $=3348-24d^2-72d$

Given, $\frac{p+q}{p-q}=\frac{19}{5}$

Thus, $\frac{3348-24d^2-72d}{72d+108}=\frac{19}{5}$

Divide numerator and denominator by $12$: $\frac{279-2d^2-6d}{6d+9}=\frac{19}{5}$

Cross-multiply: $5(279-2d^2-6d)=19(6d+9)$

$1395-10d^2-30d=114d+171$

$1395-171=10d^2+144d$

$1224=10d^2+144d$

$5d^2+72d-612=0$

Solve: $5d^2+102d-30d-612=0$ $d(5d-30)+34(5d-30)=0$ $(d+34)(5d-30)=0$

So, $d=-34 \quad \text{or} \quad d=6$

Since $d>0$, we take: $d=6$

Then, $D=d+3=9$

4. Compute $p-q$

Using $p-q=12(D^2-d^2)=12(9^2-6^2)=12(81-36)=12\cdot 45=540$

So, $\boxed{p-q=540}$

5. Check with options

Option A is $540$.

Hence the correct answer is: $\boxed{\text{A}}$

6. Compare with stored correct answer

Stored correct answer: A

Our derived answer: A

They agree.