Why are you assuming that $\displaystyle b_1,b_2\not|p $? Do you actually mean $\displaystyle p\not|b_1,b_2 $?
Why $\displaystyle b_1\equiv g^n,\;b_2\equiv g^m \mod{p} $? What are n and m?Then $\displaystyle b_1\equiv g^n,\;b_2\equiv g^m \mod{p} $, so $\displaystyle g^{3n}\equiv g^{3m}\mod{p}\implies 3n\equiv 3m\mod{\phi(p)} $.
On the last line, I think you're using Fermat's little theorem, but how do you know that $\displaystyle p\not|g $?Since $\displaystyle (3,p-1)=1 \implies n\equiv m\mod{\phi(p)} \implies m=n+a(p-1) $
So $\displaystyle b_2\equiv g^m=g^{n+a(p-1)} \equiv 1\cdot g^n\equiv b_1\mod{p} $.
Thanks for clarifying this!
This is my version of the definition of primitive root:
a is a primitive root mod m iff the order of a mod m is equal to phi(m) (where the order of a mod m is the smallest positive integer h such that a^h is congruent to 1 mod m.)
But from here I really don't see how we can get b1=g^n (mod p)...