First, find the derivative of the curve \(y = \frac{4}{\sqrt{x}}\). Rewrite it as \(y = 4x^{-0.5}\).

The derivative is \(\frac{dy}{dx} = -2x^{-1.5}\).

At the point \((4, 2)\), the gradient of the tangent is \(\frac{dy}{dx} = -\frac{1}{4}\).

The gradient of the normal is the negative reciprocal, so \(m = 4\).

The equation of the normal is \(y - 2 = 4(x - 4)\).

Simplify to get \(y = 4x - 14\).

Find where the normal meets the \(x\)-axis by setting \(y = 0\):

\(0 = 4x - 14\)

\(x = 3.5\)

So, \(P(3.5, 0)\).

Find where the normal meets the \(y\)-axis by setting \(x = 0\):

\(y = -14\)

So, \(Q(0, -14)\).

The length of \(PQ\) is given by:

\(\sqrt{(3.5 - 0)^2 + (0 + 14)^2} = \sqrt{3.5^2 + 14^2}\)

\(= \sqrt{12.25 + 196} = \sqrt{208.25}\)

\(\approx 14.4\)