The line $2y + x + a = 0$ is a normal to the curve $y = 2x^2 - 3x + 1$ at the point $P$.

1. Find the gradient of the normal from the given line equation.

The equation of the normal is $2y + x + a = 0$.
Rearranging into the form $y = mx + c$:
$2y = -x - a$
$y = -\frac{1}{2}x - \frac{a}{2}$
The gradient of the normal is $-\frac{1}{2}$.

2. Determine the gradient of the tangent at $P$.

Let the gradient of the tangent be $m_t$.
Since the normal is perpendicular to the tangent, $m_t \times \left(-\frac{1}{2}\right) = -1$.
$m_t = 2$

3. Find the derivative of the curve and equate it to the gradient of the tangent to find the $x$-coordinate of $P$.

The equation of the curve is $y = 2x^2 - 3x + 1$.
$\frac{dy}{dx} = 4x - 3$
At point $P$, the gradient of the tangent is $2$:
$4x - 3 = 2$
$4x = 5$
$x = \frac{5}{4}$

4. Substitute the $x$-coordinate into the curve equation to find the $y$-coordinate of $P$.

Substitute $x = \frac{5}{4}$ into $y = 2x^2 - 3x + 1$:
$y = 2\left(\frac{5}{4}\right)^2 - 3\left(\frac{5}{4}\right) + 1$
$y = 2\left(\frac{25}{16}\right) - \frac{15}{4} + 1$
$y = \frac{25}{8} - \frac{30}{8} + \frac{8}{8}$
$y = \frac{3}{8}$
The coordinates of $P$ are $\left(\frac{5}{4}, \frac{3}{8}\right)$.

1. Find the coordinates of $P$ (from part i) by equating the derivative to the gradient of the tangent

The normal line is $2y + x + a = 0$, which has a gradient of $-\frac{1}{2}$.
The gradient of the tangent at $P$ is therefore $2$.
$\frac{dy}{dx} = 4x - 3$
$4x - 3 = 2 \implies x = \frac{5}{4}$
$y = 2\left(\frac{5}{4}\right)^2 - 3\left(\frac{5}{4}\right) + 1 = \frac{3}{8}$
So, $P$ is $\left(\frac{5}{4}, \frac{3}{8}\right)$.

2. Substitute the coordinates of $P$ into the equation of the normal to find $a$

Since $P$ lies on the normal $2y + x + a = 0$:
$2\left(\frac{3}{8}\right) + \frac{5}{4} + a = 0$
$\frac{3}{4} + \frac{5}{4} + a = 0$
$\frac{8}{4} + a = 0$
$2 + a = 0 \implies a = -2$