1. **State the problem:** Simplify the expression $$\frac{9\sqrt{432p^5} - 7\sqrt{27p^5}}{\sqrt{33p^4}}, p > 0.$$\n\n2. **Recall the formula and rules:** \n- The square root of a product is the product of the square roots: $$\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}.$$\n- For variables with exponents under a square root: $$\sqrt{p^n} = p^{\frac{n}{2}}.$$\n- Since $$p > 0$$, we can simplify without absolute values.\n\n3. **Simplify each square root in the numerator:**\n- $$\sqrt{432p^5} = \sqrt{432} \cdot \sqrt{p^5}.$$\n- Factor 432: $$432 = 16 \times 27,$$ so $$\sqrt{432} = \sqrt{16 \times 27} = \sqrt{16} \cdot \sqrt{27} = 4 \cdot 3\sqrt{3} = 12\sqrt{3}.$$\n- Simplify $$\sqrt{p^5} = p^{\frac{5}{2}} = p^2 \cdot p^{\frac{1}{2}} = p^2 \sqrt{p}.$$\n- Therefore, $$\sqrt{432p^5} = 12\sqrt{3} \cdot p^2 \sqrt{p} = 12 p^2 \sqrt{3p}.$$\n\n- Similarly, $$\sqrt{27p^5} = \sqrt{27} \cdot \sqrt{p^5} = 3\sqrt{3} \cdot p^2 \sqrt{p} = 3 p^2 \sqrt{3p}.$$\n\n4. **Rewrite numerator:**\n$$9\sqrt{432p^5} - 7\sqrt{27p^5} = 9 \times 12 p^2 \sqrt{3p} - 7 \times 3 p^2 \sqrt{3p} = 108 p^2 \sqrt{3p} - 21 p^2 \sqrt{3p} = (108 - 21) p^2 \sqrt{3p} = 87 p^2 \sqrt{3p}.$$\n\n5. **Simplify denominator:**\n$$\sqrt{33p^4} = \sqrt{33} \cdot \sqrt{p^4} = \sqrt{33} \cdot p^2.$$\n\n6. **Rewrite the entire expression:**\n$$\frac{87 p^2 \sqrt{3p}}{p^2 \sqrt{33}}.$$\n\n7. **Cancel common factors:**\n$$\frac{\cancel{p^2} 87 \sqrt{3p}}{\cancel{p^2} \sqrt{33}} = 87 \frac{\sqrt{3p}}{\sqrt{33}} = 87 \sqrt{\frac{3p}{33}} = 87 \sqrt{\frac{p}{11}}.$$\n\n8. **Final simplified form:**\n$$\boxed{87 \sqrt{\frac{p}{11}}}.$$