To find \(f^{-1}(x)\), start by setting \(y = f(x) = 2 - \frac{3}{4x - p}\).

Rearrange to solve for \(x\):

\(y - 2 = -\frac{3}{4x - p}\)

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

\(4xy - 8x = py - 2p - 3\)

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

\(4x = \frac{p(y - 2) - 3}{y - 2}\)

\(x = \frac{p(y - 2) - 3}{4(y - 2)}\)

\(x = \frac{p}{4} - \frac{3}{4(y - 2)}\)

Thus, \(f^{-1}(x) = \frac{p}{4} - \frac{3}{4x - 8}\).

For \(f^{-1}(x) \equiv f(x)\), equate the expressions:

\(2 - \frac{3}{4x - p} = \frac{p}{4} - \frac{3}{4x - 8}\)

Equating the coefficients gives \(p = 8\).