We claim the left side is the coefficient of $x^p$ in the expanded form of $(1+x)^p(2+x)^p$.

This is because $\binom{p+j}{j} = \binom{p+j}{p}$, therefore the left side is the coefficient of $x^p$ in the following polynomial $$\begin{align*} \sum_{j=0}^p\left(\binom{p}{j}\displaystyle\sum_{k=0}^{p+j}\binom{p+j}{k}\cdot x^k\right)=\ &\sum_{j=0}^p\binom{p}{j}(1+x)^{p+j} \\=\ &(1+x)^p\sum_{j=0}^p\binom{p}{j}(1+x)^{j}\\ =&(1+x)^p(2+x)^p \end{align*}$$

On the other hand, expanding directly shows the coefficient of $x^p$ is $$\sum_{k=0}^{p}\binom{p}{k}\binom{p}{p-k}\cdot 2^k=\sum_{k=0}^{p}\left(\frac{p!}{k!(p-k)!}\right)^2\cdot 2^k$$

The conclusion of # 4232 states $p\mid \binom{n}{k}$ when $0 < k < p$ which means $p^2\mid\binom{n}{k}^2=\left(\frac{p!}{k!(p-k)!}\right)^2$. Hence, all the terms in the above expressions are multiples of $p^2$, except when $k=0$ and $k=p$. Therefore, the original expression's modulo $p^2$ equals $(1+2^p)$.