Let $(q;q)_n=(1-q)(1-q^2)\cdots(1-q^n)$ with $(q;q)_0:=1$. Define a $q$-exponential by $$e(z;q)=\sum_{n\geq0}\frac{z^n}{(q;q)_n}.$$ There is a notion of $q$-Eulerian polynomials, see [the reference][1]. I like to introduce $q$-Eulerian polynomial of type B via the generating function $$\sum_{n\geq1}B_n(t,q)\frac{z^n}{(q;q)_n}=\frac{(e(z;q)-e(tz;q))(e(tz;q)+te(z;q))}{e(2tz;q)-te(2z;q)}.$$ Now, expand $B_n(t,q)$ as a polynomial $$B_n(t,q)=\sum_{k=0}^nB_{n,k}(q)t^k$$ and call $B_{n,k}(q)$ $q$-Eulerian numbers type B.

Claim. if $a, b\in\Bbb{N}$ and $\alpha=a+b+1$, then the symmetric relation holds: $$\binom{\alpha}a_q+\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,b}(q)= \binom{\alpha}b_q +\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,a}(q).$$

QUESTIONS:

(a) I don't have a proof for my claim which seems very true though. Do you?

(b) Is there a combinatorial interpretation for these polynomials $B_n(t,q)$ or the Eulerian numbers $B_{n,k}(q)$? You might be inspired by [the reference][1]. [1]: https://arxiv.org/pdf/1201.4941.pdf