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$\newcommand\si\sigma$This is not true in general. Indeed, $$B(n_1)=B_\si(n_1):=E(((1-t)N_2+t(N-n_1))^\si|N_1=n_1)$$ for random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$.

If the inequality in question were true for all real $\si>1$, then we would have $$d_t(n_1):=M_\infty(n_1)-M_\infty(n_1+1)\le1 \tag{1}\label{1}$$ for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, where $$M_\infty(n_1):=\lim_{\si\to\infty}B_\si(n_1)^{1/\si}=(1-t)m(n_1)+t(N-n_1), $$ where in turn $m(n_1)$ is the maximum of the support set of the conditional distribution of $N_2$ given $N_1=n_1$. So, $$d_t(n_1)=(1-t)(m(n_1)-m(n_1+1))+t.$$ So, if \eqref{1} were true for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, then we would have $$d_0(n_1)=m(n_1)-m(n_1+1)\le1 \tag{2}\label{2}$$ for all $n_1\in\{0,\dots,N-1\}$.

However, it is easy to construct random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$ while \eqref{2} fails to hold for some $n_1\in\{0,\dots,N-1\}$.

E.g., let $N=3$, $p=1/2$, $P(N_2=1|N_1=1)=1$, and $P(N_2=n_2|N_1=n_1)=\frac1{4-n_1}$ for $n_1\in\{0,2,3\}$ and $n_2\in\{0,\dots,3-n_1\}$. Then all the conditions on $N_1$ and $N_2$ will hold, while $d_0(0)=m(0)-m(1)=3-1=2\not\le1$, so that \eqref{2} fails to hold.

Actually, for these random variables $N_1$ and $N_2$, $t=0$, and any $\si>1$ we have $$B_\si(0)^{1/\si}-B_\si(1)^{1/\si} =4^{-1/\sigma } \left(3^{\sigma }+2^{\sigma }+1\right)^{1/\si}-1>1$$ already for $\si\ge2.57$. $\quad\Box$

$\newcommand\si\sigma$This is not true in general. Indeed, $$B(n_1)=B_\si(n_1):=E(((1-t)N_2+t(N-n_1))^\si|N_1=n_1)$$ for random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$.

If the inequality in question were true for all real $\si>1$, then we would have $$d_t(n_1):=M_\infty(n_1)-M_\infty(n_1+1)\le1 \tag{1}\label{1}$$ for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, where $$M_\infty(n_1):=\lim_{\si\to\infty}B_\si(n_1)^{1/\si}=(1-t)m(n_1)+t(N-n_1), $$ where in turn $m(n_1)$ is the maximum of the support set of the conditional distribution of $N_2$ given $N_1=n_1$. So, $$d_t(n_1)=(1-t)(m(n_1)-m(n_1+1))+t.$$ So, if \eqref{1} were true for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, then we would have $$d_0(n_1)=m(n_1)-m(n_1+1)\le1 \tag{2}\label{2}$$ for all $n_1\in\{0,\dots,N-1\}$.

However, it is easy to construct random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$ while \eqref{2} fails to hold for some $n_1\in\{0,\dots,N-1\}$.

E.g., let $N=3$, $p=1/2$, $P(N_2=1|N_1=1)=1$, and $P(N_2=n_2|N_1=n_1)=\frac1{4-n_1}$ for $n_1\in\{0,2,3\}$ and $n_2\in\{0,\dots,3-n_1\}$. Then all the conditions on $N_1$ and $N_2$ will hold, while $d_0(0)=m(0)-m(1)=3-1=2\not\le1$, so that \eqref{2} fails to hold.

Actually, for these random variables $N_1$ and $N_2$, $t=0$, and any $\si>1$ we have $$D(\si):=B_\si(0)^{1/\si}-B_\si(1)^{1/\si} =4^{-1/\sigma } \left(3^{\sigma }+2^{\sigma }+1\right)^{1/\si}-1>1$$ already for $\si\ge2.57$. $\quad\Box$

For an illustration, here is the graph $\{(\si,D(\si))\colon1<\si<12\}$ (solid):

$\newcommand\si\sigma$This is not true in general. Indeed, $$B(n_1)=B_\si(n_1):=E(((1-t)N_2+t(N-n_1))^\si|N_1=n_1)$$ for random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$.

If the inequality in question were true for all real $\si>1$, then we would have $$d_t(n_1):=M_\infty(n_1)-M_\infty(n_1+1)\le1 \tag{1}\label{1}$$ for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, where $$M_\infty(n_1):=\lim_{\si\to\infty}B_\si(n_1)^{1/\si}=(1-t)m(n_1)+t(N-n_1), $$ where in turn $m(n_1)$ is the maximum of the support set of the conditional distribution of $N_2$ given $N_1=n_1$. So, $$d_t(n_1)=(1-t)(m(n_1)-m(n_1+1))+t.$$ So, if \eqref{1} were true for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, then we would have $$d_0(n_1)=m(n_1)-m(n_1+1)\le1 \tag{2}\label{2}$$ for all $n_1\in\{0,\dots,N-1\}$.

However, it is easy to construct random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$ while \eqref{2} fails to hold for some $n_1\in\{0,\dots,N-1\}$.

E.g., let $N=3$, $p=1/2$, $P(N_2=1|N_1=1)=1$, and $P(N_2=n_2|N_1=n_1)=\frac1{4-n_1}$ for $n_1\in\{0,2,3\}$ and $n_2\in\{0,\dots,3-n_1\}$. Then all the conditions on $N_1$ and $N_2$ will hold, while $d_0(0)=m(0)-m(1)=3-1=2\not\le1$, so that \eqref{2} fails to hold. $\quad\Box$

$\newcommand\si\sigma$This is not true in general. Indeed, $$B(n_1)=B_\si(n_1):=E(((1-t)N_2+t(N-n_1))^\si|N_1=n_1)$$ for random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$.

If the inequality in question were true for all real $\si>1$, then we would have $$d_t(n_1):=M_\infty(n_1)-M_\infty(n_1+1)\le1 \tag{1}\label{1}$$ for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, where $$M_\infty(n_1):=\lim_{\si\to\infty}B_\si(n_1)^{1/\si}=(1-t)m(n_1)+t(N-n_1), $$ where in turn $m(n_1)$ is the maximum of the support set of the conditional distribution of $N_2$ given $N_1=n_1$. So, $$d_t(n_1)=(1-t)(m(n_1)-m(n_1+1))+t.$$ So, if \eqref{1} were true for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, then we would have $$d_0(n_1)=m(n_1)-m(n_1+1)\le1 \tag{2}\label{2}$$ for all $n_1\in\{0,\dots,N-1\}$.

However, it is easy to construct random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$ while \eqref{2} fails to hold for some $n_1\in\{0,\dots,N-1\}$.

E.g., let $N=3$, $p=1/2$, $P(N_2=1|N_1=1)=1$, and $P(N_2=n_2|N_1=n_1)=\frac1{4-n_1}$ for $n_1\in\{0,2,3\}$ and $n_2\in\{0,\dots,3-n_1\}$. Then all the conditions on $N_1$ and $N_2$ will hold, while $d_0(0)=m(0)-m(1)=3-1=2\not\le1$, so that \eqref{2} fails to hold.

Actually, for these random variables $N_1$ and $N_2$, $t=0$, and any $\si>1$ we have $$B_\si(0)^{1/\si}-B_\si(1)^{1/\si} =4^{-1/\sigma } \left(3^{\sigma }+2^{\sigma }+1\right)^{1/\si}-1>1$$ already for $\si\ge2.57$. $\quad\Box$

$\newcommand\si\sigma$This is not true in general. Indeed, $$B(n_1)=B_\si(n_1):=E(((1-t)N_2+t(N-n_1))^\si|N_1=n_1)$$ for random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$.

If the inequality in question were true for all real $\si>1$, then we would have $$d_t(n_1):=M_\infty(n_1)-M_\infty(n_1+1)\le1 \tag{1}\label{1}$$ for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, where $$M_\infty(n_1):=\lim_{\si\to\infty}B_\si(n_1)^{1/\si}=(1-t)m(n_1)+t(N-n_1), $$ where in turn $m(n_1)$ is the maximum of the support set of the conditional distribution of $N_2$ given $N_1=n_1$. So, $$d_t(n_1)=(1-t)(m(n_1)-m(n_1+1))+t.$$ So, if \eqref{1} were true for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, then we would have $$d_0(n_1)=m(n_1)-m(n_1+1)\le1 \tag{2}\label{2}$$ for all $n_1\in\{0,\dots,N-1\}$. However, it is easy to construct random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$ such that \eqref{2} fails to hold for some $n_1\in\{0,\dots,N-1\}$. $\quad\Box$

$\newcommand\si\sigma$This is not true in general. Indeed, $$B(n_1)=B_\si(n_1):=E(((1-t)N_2+t(N-n_1))^\si|N_1=n_1)$$ for random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$.

If the inequality in question were true for all real $\si>1$, then we would have $$d_t(n_1):=M_\infty(n_1)-M_\infty(n_1+1)\le1 \tag{1}\label{1}$$ for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, where $$M_\infty(n_1):=\lim_{\si\to\infty}B_\si(n_1)^{1/\si}=(1-t)m(n_1)+t(N-n_1), $$ where in turn $m(n_1)$ is the maximum of the support set of the conditional distribution of $N_2$ given $N_1=n_1$. So, $$d_t(n_1)=(1-t)(m(n_1)-m(n_1+1))+t.$$ So, if \eqref{1} were true for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, then we would have $$d_0(n_1)=m(n_1)-m(n_1+1)\le1 \tag{2}\label{2}$$ for all $n_1\in\{0,\dots,N-1\}$.

However, it is easy to construct random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$ while \eqref{2} fails to hold for some $n_1\in\{0,\dots,N-1\}$.

E.g., let $N=3$, $p=1/2$, $P(N_2=1|N_1=1)=1$, and $P(N_2=n_2|N_1=n_1)=\frac1{4-n_1}$ for $n_1\in\{0,2,3\}$ and $n_2\in\{0,\dots,3-n_1\}$. Then all the conditions on $N_1$ and $N_2$ will hold, while $d_0(0)=m(0)-m(1)=3-1=2\not\le1$, so that \eqref{2} fails to hold. $\quad\Box$