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. 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**. Here are the first few terms:
\begin{align} B_1(t,q)&=1+t, \\
B_2(t,q)&=1+(2q+4)t+t^2, \\
B_3(t,q)&=1+(7q^2+7q+9)t+(7q^2+7q+9)t^2+t^3.
\end{align}

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.