The coefficient of \(x^4\) in the expansion of \((x + a)^6\) is given by \(\binom{6}{4} a^2 = 15a^2\). Thus, \(p = 15a^2\).

The coefficient of \(x^2\) in the expansion of \((ax + 3)^4\) is given by \(\binom{4}{2} (ax)^2 3^2 = 54a^2\). Thus, \(q = 54a^2\).

We know that \(p + q = 276\), so:

\(15a^2 + 54a^2 = 276\)

\(69a^2 = 276\)

\(a^2 = \frac{276}{69} = 4\)

\(a = \pm 2\)