Solution:

a.
Let $Q_0$ be the original quadruple $(a, b, c, d)$ and $Q_n$ the quadruple after $n$ transformations. If $abcd > 1$, then the products form a strictly increasing sequence, so return is impossible. Similarly if $abcd < 1$. So we must have $abcd = 1$. Let the largest of the four values of a quadruple $Q$ be $M(Q)$. If a member of $Q_1$ is not $1$, then $M(Q_1) > 1$. $Q_3$ consists of the elements of $Q_1$ squared and permuted, so $M(Q_3) = M(Q_1)^2$. Hence the sequence $M(Q_1)$, $M(Q_3)$, $M(Q_5)$, ... increases without limit. This means no return is possible, because a return would lead to the values cycling.

b.
After $r < n$ transformations, the first number of the $n$-tuple is the product $a_1^{(r\choose 0)} a_2^{(r\choose 1)} \ldots a_{r+1}^{(r\choose r)}$, where $(r\choose i)$ denotes the binomial coefficient. [This is an easy induction.] Hence after $n = 2^k$ transformations it is $a_1^2$ times the product $a_2^{(n\choose 1)} \ldots a_n^{(n\choose 1)}$. So it is sufficient to prove that $(n\choose i)$ is even for $n$ a power of $2$ and $0 < i < n$. But observe that $\text{(n i) = (n-1 i) n/(n-i)}$ and $n$ is divisible by a higher power of $2$ than $n-i$.