Yes we can. The question is deceptive, because even weaker hypotheses actually imply a much stronger conclusion: that $M$ is normal in $N_G(P)$. The argument is valid for any nilpotent subgroup $P$ of $G$ and any subgroup $M\le P$ such that $MM^h=M^hM$ implies $M=M^h$ whenever $h\in N_G(P)$.

Proof: First, let $R=N_P(N_P(M))$. Now $M\le N_P(M)$. Therefore for any $h\in R$, $M^h\le N_P(M)$, so $MM^h=M^hM$. Hence by assumption $h\in N_P(M)$. Thus $N_P(M)=R$ is its own normalizer in $P$, whence $N_P(M)=P$ by nilpotence of $P$. Hence, $M$ is normal in $P$. Finally, for any $h\in N_G(P)$, $M^h\le P$ so again $MM^h=M^hM$, whence $h\in N_G(M)$. Thus, $M$ is normal in $N_G(P)$.

It is then elementary that $MN$ is normal in $N_G(P)N$, which implies your conclusion.

Yes we can. The question is deceptive, because even weaker hypotheses actually imply a much stronger conclusion: that $M$ is normal in $N_G(P)$. The argument is valid for any nilpotent subgroup $P$ of $G$ and any subgroup $M\le P$ such that for any $h\in N_G(P)$, $MM^h=M^hM$ implies $M=M^h$.

Proof: First, let $R=N_P(N_P(M))$. Now $M\le N_P(M)$. Therefore for any $h\in R$, $M^h\le N_P(M)$, so $MM^h=M^hM$. Hence by assumption $h\in N_P(M)$. Thus $N_P(M)=R$ is its own normalizer in $P$, whence $N_P(M)=P$ by nilpotence of $P$. Hence, $M$ is normal in $P$. Finally, for any $h\in N_G(P)$, $M^h\le P$ so again $MM^h=M^hM$, whence $h\in N_G(M)$. Thus, $M$ is normal in $N_G(P)$.

It is then elementary that $MN$ is normal in $N_G(P)N$, which implies your conclusion.

Yes we can. The question is deceptive, because the hypotheses actually imply a much stronger conclusion: that $M$ is normal in $N_G(P)$.

Proof: First, let $R=N_P(N_P(M))$. Now $M\le N_P(M)$. Therefore for any $h\in R$, $M^h\le N_P(M)$, so $MM^h=M^hM$. Hence by assumption $h\in N_P(M)$. Thus $N_P(M)=R$ is its own normalizer in $P$, whence $N_P(M)=P$ by nilpotence of $P$. Hence, $M$ is normal in $P$. Finally, for any $h\in N_G(P)$, $M^h\le P$ so again $MM^h=M^hM$, whence $h\in N_G(M)$. Thus, $M$ is normal in $N_G(P)$.

It is then elementary that $MN$ is normal in $N_G(P)N$, which implies your conclusion.

Yes we can. The question is deceptive, because even weaker hypotheses actually imply a much stronger conclusion: that $M$ is normal in $N_G(P)$. The argument is valid for any nilpotent subgroup $P$ of $G$ and any subgroup $M\le P$ such that $MM^h=M^hM$ implies $M=M^h$ whenever $h\in N_G(P)$.

Proof: First, let $R=N_P(N_P(M))$. Now $M\le N_P(M)$. Therefore for any $h\in R$, $M^h\le N_P(M)$, so $MM^h=M^hM$. Hence by assumption $h\in N_P(M)$. Thus $N_P(M)=R$ is its own normalizer in $P$, whence $N_P(M)=P$ by nilpotence of $P$. Hence, $M$ is normal in $P$. Finally, for any $h\in N_G(P)$, $M^h\le P$ so again $MM^h=M^hM$, whence $h\in N_G(M)$. Thus, $M$ is normal in $N_G(P)$.

It is then elementary that $MN$ is normal in $N_G(P)N$, which implies your conclusion.