In my question I already verified that (assuming 1. and 2.) item 3. is provable without assuming that $\pi(r)$ is $k$-solid over $(\mathcal{M},q)$. To see that we can actually drop $k$-solidity in this lemma, it hence suffices to see that 1. and 2. also don't require that $k$-solidity of $\pi(r)$.

If $\rho_{k}^{\mathcal{M}} = \operatorname{Ord}^{\mathcal{M}}$, then $\pi = \operatorname{id}$
  and the lemma trivially holds. Thus assume that
  $\rho_{k}^{\mathcal{M}} < \operatorname{Ord}^{\mathcal{M}}$.

 1. Let $\alpha \leq \rho_{k}^{\mathcal M}$. Since
    $\pi \restriction \rho_{\kappa}^{\mathcal{M}} = \operatorname{id}$
    and since $\pi$ is generalized $r \Sigma_{k}$-elementary, we have - up
    to a slight abuse of notation -
    \begin{align*}
      \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \})
      &= \{(\phi, \vec{a}, s) \mid \vec{a} \in ^{< \omega}{\alpha}
        \wedge \mathcal{H} \models \phi[\vec{a}, s] \} \\
      &= \{(\phi, \vec{a}, s) \mid \vec{a} \in ^{< \omega}{\alpha}
        \wedge \mathcal{M} \models \phi[\vec{a}, \pi(s)] \}
    \end{align*}
    In particular, for any $\phi \in r \Sigma_{k}$ and any $\vec{a} \in ^{<
      \omega}{\alpha}$, we have
    $$
      (\phi, \vec{a}, s) \in
      \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{s \}) \iff
      (\phi, \vec{a}, \pi(s)) \in
      \operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}).
    $$
    By enlarging $\alpha$, if necessary, we may assume that $\alpha$
    is primitive recursively closed and hence uniformly code
    \begin{align*}
      \{(\phi, \vec{a}) \mid (\phi, \vec{a}, s) \in
      \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) \} \\
      = \{(\phi, \vec{a}) \mid (\phi, \vec{a}, \pi(s)) \in
      \operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}) \}
    \end{align*}
    as a subset $A \subseteq \alpha$. Since
    $\alpha < \rho_{k}^{\mathcal{M}}$, we have
    $\operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}) \in
    \mathcal{M}$ and hence $A \in \mathcal{M}$. By the strong
    acceptability of $\mathcal{M}$ - observing that
    $\rho_{k}^{\mathcal{M}}$ is an $\mathcal{M}$-cardinal - this
    yields
    $$
      A \in \mathcal{J}_{\rho_{k}^{\mathcal{M}}}^{\mathcal M} = \left(
        H_{\rho_{k}^{\mathcal{M}}} \right)^{\mathcal{M}} \overset{\pi
        \restriction \rho_{k}^{\mathcal{M}} = \operatorname{id}}{=} \left(
        H_{\rho_{k}^{\mathcal{M}}}\right)^{\mathcal{H}} \subseteq \mathcal{H}.
    $$
    Therefore
    $$
      \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) = \{
      (\phi, \vec{a}, s) \mid \langle \phi, \vec{a} \rangle \in A \}
      \in \mathcal{H}
    $$
    and $\rho_{k}^{\mathcal{M}} \leq \rho_{k}^{\mathcal{H}}$.


    On the other hand, suppose that
    $\rho_{k}^{\mathcal{M}} < \rho_{k}^{\mathcal{H}}$. Then
    $\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{M}} \cup
    \{(r,q)\}) \in \mathcal{H}$. Let
    $A \subseteq \rho_{k}^{\mathcal{M}}$, $A \in \mathcal{H}$ be a
    uniform code for
    $$
      \{ (\phi, \vec{a}) \mid (\phi, \vec{a}, (r,q)) \in
      \operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{M}} \cup
      \{(r,q)\})\}.
    $$
    Then $A = \pi(A) \cap \rho_{k}^{\mathcal{M}} \in \mathcal{M}$
    witnesses (as above) that
    $\operatorname{Th}_{k}^{\mathcal{M}}(\rho_{k}^{\mathcal{M}} \cup
    \{ \pi(r), \pi(q)\}) \in \mathcal{M}$. This contradicts the fact
    that $\pi(r)$ is the $k$th standard parameter of
    $(\mathcal{M},\pi(q))$!
 2. The proof above also shows that
    $\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup
    \{ (r,q)\}) \not \in \mathcal{H}$. Hence it suffices to show that
    for all $s <_{\operatorname{lex}} r$
    $$
      \operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup
      \{ (s,q)\}) \in \mathcal{H}.
    $$
    So, fix $s <_{\operatorname{lex}} r$. Then
    $\pi(s)<_{\operatorname{lex}} \pi(r)$ and hence
    $$
      \operatorname{Th}_{k}^{\mathcal{M}}(\rho_{k}^{\mathcal{H}} \cup
      \{ (\pi(s),\pi(q))\}) \in \mathcal{M}.
    $$
    Let $A \subseteq \rho_{k}^{\mathcal{M}}$, $A \in \mathcal{M}$ be
    the code of this fact as above. Since $\pi(r)$ is $k$-universal
    over $(M, \pi(q))$ there is some $\tau \in \operatorname{Sk}_{k}$
    and $\vec{a} \in ^{< \omega}{\rho_{k}^{\mathcal{M}}}$ such that
    $A = \tau^{\mathcal{M}}[\vec{a}, \pi(r), \pi(q)] \cap
    \rho_{k}^{\mathcal{M}}$.  Let
    $$
      B = \tau^{\mathcal{H}}[\vec{a}, \pi(r), \pi(q)] \cap
      \rho_{k}^{\mathcal{M}}.
    $$
    Since $\pi$ is *generalized* $r \Sigma_{k}$-elementary we have,
    for all $\xi < \rho_{k}^{\mathcal{M}} = \rho_{k}^{\mathcal{H}}$
    $$
      \mathcal{M} \models \xi \in
      \tau^{\mathcal{M}}[\vec{a},\pi(r),\pi(q)] \iff \mathcal{H}
      \models \xi \in \tau^{\mathcal{H}}[\vec{a},r,q].
    $$
    Thus $B = A \in \mathcal{H}$ witnesses that
    $\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup
    \{(s,q)\}) \in \mathcal{H}$ and hence the
    $<_{\operatorname{lex}}$-minimality of $r$.