I need help with the Hardy-Weinberg principle please.?

You're forgetting the Heterozygotes 2pq
The full formula is p^2+2pq+q^2 = 1
We know that 2/3 of the Red flowers are heterozygotes
p^2 = 28 AA Red
2pq = 56 Aa Red
q^2 = 16 aa White

So the allele frequencies p are:
2 A's from the homozygous = 28 +
1 A from the heterozygous = 56/2 = 28
= 56/100 = .56

So the allele frequencies q are:
1-.56 = .44

Always work with the recessive phenotype first.    It seems you have forgotten to take the sqrt of the known genotype, which is 16/100.  If 16/100 have white flowers, that's a frequency of 0.16.  Then, the frequency of the recessive homozygote is 0.16 (using Hardy-Weinberg notation, q?2 = 0.16).  The frequency of the recessive allele, q, is sqrt(0.16) = 0.4.  According to H-W equation #1, p + q = 1, so the frequency of the dominant allele, p, is 1 - 0.4 = 0.6.

If equilibrium is maintained, p will still be 0.6 and q will still be 0.4.

Just to finish it off:  The frequency of the dominant homozygote, AA = p?2 = 0.36;  the frequency of the heterozygotes, Aa = 2pq = 0.48, and the frequency of the recessive homozygotes, aa = q?2 = 0.16.  Add up the frequencies and you get 1.

Asst Prof is correct.  In a population, you MUST start with the recessive phenotype;  it's the only genotype you can be sure of.  In this case, the frequency of the recessive genotype, aa, is 0.16, so the frequency of a is the squareroot of 0.16, which is 0.4.  that's called q. Since p + q =1, p must be 0.6.