\[ \begin{align*} S_q &= \sum_{m=1}^{\frac{p-1}{2}} \frac{1}{(3m-1)(3m)(3m+1)} \\
&= \sum_{m=1}^{\frac{p-1}{2}} \frac{1}{2} \left( \frac{1}{3m-1} + \frac{1}{3m+1} - \frac{2}{3m} \right) \\ &= \frac12 \left( \sum_{k=2}^{q+2} \left( \frac1k \right)- 3 \sum_{m=1}^{\frac{p-1}{2}} \left( \frac{1}{3m} \right) \right) \\ &=  \frac12 \left( \sum_{k=2}^{q+2} \left( \frac1k \right)- 3 \sum_{m=1}^{\frac{p-1}{2}} \left( \frac{1}{m} \right) \right) \end{align*} \]

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\[ \begin{align*} \frac{1}{p} - 2S_q &= \sum_{m=1}^{\frac{p-1}{2}} \left( \frac{1}{m} \right) - \sum_{k=2}^{p-1} \left( \frac{1}{k} \right) - \sum_{k=p+1}^{q+2} 
\left( \frac{1}{k} \right) \\ 
&\equiv \sum_{m=1}^{\frac{p-1}{2}} \left( \frac{1}{m} \right) - \sum_{k=2}^{p-1} \left( \frac{1}{k} \right) - \sum_{k=(p+1)-p}^{q+2-p} \left(\frac{1}{k} \right) \pmod{p} \\ 
&\equiv \sum_{m=1}^{\frac{p-1}{2}} \left( \frac{1}{m} \right) - \sum_{k=2}^{p-1} \left( \frac{1}{k} \right) - \sum_{k=1}^{\frac{p-1}{2}} \left(\frac{1}{k} \right) \pmod{p} \\ 
&= - \sum_{k=2}^{p-1} k^{-1} \pmod{p} \\ 
&\equiv -(-1) \pmod{p} \\
 & \equiv 1 \pmod{p} \end{align*} \]

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