We don't even need $T(a)$. Here's a proof that $gcd(P(a),Q(a),R(a),S(a))=1$.
Fix $a\in \mathbb{N}$ and suppose $d$ is a common divisor of each of $P(a)$, $Q(a)$, $R(a)$, and $S(a)$. Since $d\vert 2a+1$, we know that $d$ is odd.
Since $Q(a)=3a+4=(a+3) + (2a+1)$, it follows that $d \vert a+3$.
Since $R(a)=4a+9=(3a+4)+(a+5)$, it follows that $d \vert a+5$.
Further, $d$ divides $(a+3) + (a+5)=2a+8=(2a+1)+7$, hence $d \vert 7\,\Rightarrow\,d\in \{1,7\}$. Suppose $d=7$.
Then $7 \vert S(a)$, hence 7 divides $(4a+9)+(a+7)$. Thus, $7 \vert a+7$ and it follows that $7 \vert a$. However, then we would have $7=d \equiv 2a+1 \equiv 1 \mod(7)$, a contradiction.
Therefore the only common divisor of $P(a)$, $Q(a)$, $R(a)$, and $S(a)$ is 1 and hence $gcd(P(a),Q(a),R(a),S(a))=1$.