1. Differentiate \(C_1: y = x^2 - 4x + 7\) to find the slope of the tangent:

\(\frac{dy}{dx} = 2x - 4\)

At \(x = 3\), the slope \(m = 2(3) - 4 = 2\).

2. The equation of the tangent at \(x = 3\) is:

\(y - 4 = 2(x - 3)\)

\(y = 2x - 2\)

3. For \(C_2: y^2 = 4x + k\), differentiate implicitly:

\(2y \frac{dy}{dx} = 4\)

\(\frac{dy}{dx} = \frac{2}{y}\)

4. Equate the slopes from \(C_1\) and \(C_2\):

\(2 = \frac{2}{y} \Rightarrow y = 1\)

5. Substitute \(y = 1\) into \(C_2\):

\(1^2 = 4x + k \Rightarrow 4x + k = 1\)

6. Solve for \(x\) using the tangent equation:

\(2x - 2 = 1 \Rightarrow x = \frac{3}{2}\)

7. Substitute \(x = \frac{3}{2}\) into \(4x + k = 1\):

\(4 \left( \frac{3}{2} \right) + k = 1 \Rightarrow 6 + k = 1 \Rightarrow k = -5\)

8. The coordinates of \(P\) are \(\left( \frac{3}{2}, 1 \right)\) or \((-1, 1)\).