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:
$$\binom{\alpha}a_q+\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,b}(q)=
\binom{\alpha}b_q +\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,a}(q).$$

I don't have a proof for my claim which seems very true though. Do you?
[1]: https://arxiv.org/pdf/1201.4941.pdf