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It's finite index by Margulis' normal subgroup theorem.

Since $H \lhd P_{2g+1}$, and then $\rho(P_{2g+1})=\Gamma(2)$ \rho(H)\lhd \rho(P_{2g+1})$. Since$\rho(P_{2g+1})$is finite index in$\rho(B_{2g+1})=\Gamma(2)$(I'm taking your word for this),then$\rho(H)\lhd \Gamma(2)$. Therefore therefore$\rho(H)$is either finite or finite-index in$\rho(P_{2g+1})$, and therefore in$\Gamma(2)$. Since you also say that$\rho(H)$is Zariski dense, it can't be finite. Since finite-index subgroups of$\Gamma(2)$has Sp_{2g}(\mathbb{Z})$ have the congruence subgroup property, I think that also means that $\rho(H)=\overline{\rho(H)}$, its congruence closure in $Sp_{2g}(\mathbb{Z})$.

2 added 3 characters in body

It's finite index by Margulis' normal subgroup theorem.

Since $H \lhd P_{2g+1}$, and $\rho(P_{2g+1})=\Gamma(2)$ (I'm taking your word for this), then $\rho(H)\lhd \Gamma(2)$. Therefore $\rho(H)$ is either finite or finite-index in $\Gamma(2)$. Since you also say that $\rho(H)$ is Zariski dense, it can't be finite. Since $\Gamma(2)$ has the congruence subgroup property, I think that also means that $\rho(H)=\overline{\rho(H)}$, its Zariski congruence closure in $Sp_{2g}(\mathbb{Z})$.

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It's finite index by Margulis' normal subgroup theorem.

Since $H \lhd P_{2g+1}$, and $\rho(P_{2g+1})=\Gamma(2)$ (I'm taking your word for this), then $\rho(H)\lhd \Gamma(2)$. Therefore $\rho(H)$ is either finite or finite-index in $\Gamma(2)$. Since you also say that $\rho(H)$ is Zariski dense, it can't be finite. Since $\Gamma(2)$ has the congruence subgroup property, I think that also means that $\rho(H)=\overline{\rho(H)}$, its Zariski closure in $Sp_{2g}(\mathbb{Z})$.