Let y = y(x) be the solution of the differential equation x...

Given the differential equation:

$x\sin\left(\dfrac{y}{x}\right)dy = \left(y\sin\left(\dfrac{y}{x}\right) - x\right)dx$

Rearranging the terms, we get:

$x\sin\left(\dfrac{y}{x}\right)dy - y\sin\left(\dfrac{y}{x}\right)dx = -xdx$

Dividing both sides by $x^2$:

$\sin\left(\dfrac{y}{x}\right)\left(\dfrac{xdy - ydx}{x^2}\right) = -\dfrac{dx}{x}$

$\sin\left(\dfrac{y}{x}\right)d\left(\dfrac{y}{x}\right) = -\dfrac{dx}{x}$

Integrating both sides:

$-\cos\left(\dfrac{y}{x}\right) = -\ln|x| - C$

$\cos\left(\dfrac{y}{x}\right) = \ln|x| + C$

Using the initial condition $y(1) = \dfrac{\pi}{2}$:

$\cos\left(\dfrac{\pi}{2}\right) = \ln(1) + C \Rightarrow C = 0$

Thus, the solution to the differential equation is:

$\cos\left(\dfrac{y}{x}\right) = \ln|x|$

We need to find $\alpha = \cos\left(\dfrac{y(e^{12})}{e^{12}}\right)$. Substituting $x = e^{12}$:

$\alpha = \ln(e^{12}) = 12$

The given equation of the circle is:

$x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$

Substituting $\alpha = 12$:

$x^2 + y^2 - 2px + 2py + 14 = 0$

The radius $r$ of the circle is given by:

$r = \sqrt{(-p)^2 + p^2 - 14} = \sqrt{2p^2 - 14}$

For the equation to represent a real circle, $r > 0$:

$2p^2 - 14 > 0 \Rightarrow p^2 > 7$

We are given that $r \leq 6$:

$\sqrt{2p^2 - 14} \leq 6$

$2p^2 - 14 \leq 36$

$2p^2 \leq 50 \Rightarrow p^2 \leq 25$

Combining the inequalities:

$7 < p^2 \leq 25$

Since $p$ is an integer, the possible values for $p^2$ are $9, 16, 25$.

This gives $p \in \{-5, -4, -3, 3, 4, 5\}$.

Therefore, there are $6$ integral values of $p$.

Answer: $6$