A curve $C$ has parametric equations
$x = 2\cos\theta + \cos 2\theta, \quad y = 2\sin\theta - \sin 2\theta,$
where $0 < \theta < 2\pi$ and $\theta \ne \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$. The point $P$ with parameter $p$ lies on $C$, and the lines $M_p$ and $N_p$ are the tangent and normal to $C$ at $P$ respectively.

1. Differentiate $x$ with respect to $\theta$ and simplify using the given trigonometric identity.

$\frac{dx}{d\theta} = -2\sin\theta - 2\sin 2\theta$
$\frac{dx}{d\theta} = -2(\sin 2\theta + \sin\theta)$
Using $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$ with $A=2\theta$ and $B=\theta$:
$\frac{dx}{d\theta} = -2 \left( 2\sin\frac{3\theta}{2}\cos\frac{\theta}{2} \right) = -4\sin\frac{3\theta}{2}\cos\frac{\theta}{2}$

2. Differentiate $y$ with respect to $\theta$ and simplify using the given trigonometric identity.

$\frac{dy}{d\theta} = 2\cos\theta - 2\cos 2\theta$
$\frac{dy}{d\theta} = 2(\cos\theta - \cos 2\theta)$
Using $\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$ with $A=\theta$ and $B=2\theta$:
$\frac{dy}{d\theta} = 2 \left( -2\sin\frac{3\theta}{2}\sin\left(-\frac{\theta}{2}\right) \right)$
Since $\sin(-\alpha) = -\sin\alpha$:
$\frac{dy}{d\theta} = 4\sin\frac{3\theta}{2}\sin\frac{\theta}{2}$

3. Use the chain rule for parametric equations to find $\frac{dy}{dx}$ and evaluate the gradient at $\theta = p$.

$\frac{dy}{dx} = \frac{\left( \frac{dy}{d\theta} \right)}{\left( \frac{dx}{d\theta} \right)}$
$\frac{dy}{dx} = \frac{4\sin\frac{3\theta}{2}\sin\frac{\theta}{2}}{-4\sin\frac{3\theta}{2}\cos\frac{\theta}{2}}$
$\frac{dy}{dx} = -\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = -\tan\left(\frac{\theta}{2}\right)$
At point $P$, $\theta = p$, so the gradient of the tangent $M_p$ is $-\tan\left(\frac{p}{2}\right)$ (shown).

1. Calculate the coordinates of point $P$ at $p = \frac{\pi}{3}$.

$x_P = 2\cos\left(\frac{\pi}{3}\right) + \cos\left(\frac{2\pi}{3}\right) = 2\left(\frac{1}{2}\right) - \frac{1}{2} = \frac{1}{2}$

$y_P = 2\sin\left(\frac{\pi}{3}\right) - \sin\left(\frac{2\pi}{3}\right) = 2\left(\frac{\sqrt{3}}{2}\right) - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

So, $P$ is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

2. Find the gradients of the tangent $M_p$ and normal $N_p$ at $P$.

From part (a), the gradient of the tangent $M_p$ is $-\tan\left(\frac{p}{2}\right)$.
At $p = \frac{\pi}{3}$, the gradient of $M_p$ is:
$m_t = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$

Since the normal $N_p$ is perpendicular to $M_p$, the gradient of $N_p$ is:
$m_n = -\frac{1}{m_t} = \sqrt{3}$

3. Determine the coordinates of $Q$ and $R$ by finding the $x$-intercepts of $M_p$ and $N_p$.

Equation of $M_p$:
$y - \frac{\sqrt{3}}{2} = -\frac{1}{\sqrt{3}}\left(x - \frac{1}{2}\right)$
To find $Q$, substitute $y = 0$:
$-\frac{\sqrt{3}}{2} = -\frac{1}{\sqrt{3}}\left(x - \frac{1}{2}\right) \implies \frac{3}{2} = x - \frac{1}{2} \implies x = 2$
So, $Q$ is $(2, 0)$.

Equation of $N_p$:
$y - \frac{\sqrt{3}}{2} = \sqrt{3}\left(x - \frac{1}{2}\right)$
To find $R$, substitute $y = 0$:
$-\frac{\sqrt{3}}{2} = \sqrt{3}\left(x - \frac{1}{2}\right) \implies -\frac{1}{2} = x - \frac{1}{2} \implies x = 0$
So, $R$ is $(0, 0)$.

4. Calculate the area of $\Delta PQR$.

The base $QR$ lies on the $x$-axis with length $2 - 0 = 2$.
The perpendicular height of $\Delta PQR$ is the $y$-coordinate of $P$, which is $\frac{\sqrt{3}}{2}$.

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2}(2)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}$

1. Equate the gradients of $M_p$ and $N_s$.

From part (a), the gradient of the tangent at parameter $\theta$ is $-\tan\left(\frac{\theta}{2}\right)$.

The gradient of the tangent $M_p$ at parameter $p$ is $m_{M_p} = -\tan\left(\frac{p}{2}\right)$.

The gradient of the normal $N_s$ at parameter $s$ is $m_{N_s} = \frac{-1}{-\tan\left(\frac{s}{2}\right)} = \frac{1}{\tan\left(\frac{s}{2}\right)}$.

Since $M_p$ and $N_s$ are parallel, their gradients are equal:
$-\tan\left(\frac{p}{2}\right) = \frac{1}{\tan\left(\frac{s}{2}\right)}$
$\tan\left(\frac{s}{2}\right) \tan\left(\frac{p}{2}\right) = -1$
$1 + \tan\left(\frac{s}{2}\right) \tan\left(\frac{p}{2}\right) = 0$

2. Apply the given trigonometric identity to find the relationship between $p$ and $s$.

Substitute $A = \frac{s}{2}$ and $B = \frac{p}{2}$ into the given identity $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$\tan\left(\frac{s}{2} - \frac{p}{2}\right) = \frac{\tan\left(\frac{s}{2}\right) - \tan\left(\frac{p}{2}\right)}{1 + \tan\left(\frac{s}{2}\right) \tan\left(\frac{p}{2}\right)}$

Since $1 + \tan\left(\frac{s}{2}\right) \tan\left(\frac{p}{2}\right) = 0$, the denominator is zero, meaning $\tan\left(\frac{s-p}{2}\right)$ is undefined.

For the tangent function to be undefined, the angle must be $\frac{\pi}{2} + k\pi$ for some integer $k$.
Given $0 < p < 2\pi$ and $0 < s < 2\pi$ with $p < s$, we know that $0 < s - p < 2\pi$, which means $0 < \frac{s-p}{2} < \pi$.

The only value in this range where tangent is undefined is $\frac{\pi}{2}$.
$\frac{s-p}{2} = \frac{\pi}{2}$
$s - p = \pi$