Public-key generation - primes reuse Given the following Trap-door Commitment scheme 


Secret key receiver: $x_B \in_u Z_q$
Public key receiver: $y_B = g^{x_B} \mod p$


Here, $p=q*k+1$ for two primes $p,q$ and $k \in Z$. And $g$ is the generator of subgrup $G_q$ of $Z_p^{*}$, of order $q$.

To the sender commit $w \in Z_q$, he selects $r \in_u Z_q$, the commitment is $c=g^wy_B^r \mod p$. To open a commitment, he sends $(w,r)$. 

My question is: if I want create milions of theses public-secret keys for milions of different senders, could I reuse $p,q$?
Yes you can reuse $p$ and $q$, i.e. you can have multiple key pairs using the same group, as long as the secrect keys $x_B$'s are chosen independently at random.

Essentially, what you are asking is this, given $y_B=g^{x_B} \in G$ where $G$ is the order-$q$ subgroup of $Z_p^*$ and $x_B \in Z_q$, and additional $y_i =g^{x_i} \in G$ for polynomial many $y_i$'s, is the trapdoor commitment is still secure?

Some more details:

The security of commitment requires two things: hiding and binding.

For the hiding property, note that your commitment scheme is perfectly (information theoretically) hiding, thus a commitment $g^w{y_B}^r$ is uniformly random and the uniformity comes from the fact $r$ is uniformly random. This also means the distribution of $g^w{y_B}^r$ is independent of all $y_i$'s the adversary can get, and those $y_i$'s provide no advantage for the adversary to learn $w$. So the hiding property holds.

For the binding property, you can reduce it to the DL problem such that if an adversary $A$ can break the binding property of your scheme, then we can use it to break the DL problem. Suppose a DL adversary who has $y=g^x \in G$ and needs to find $x$, he can choose random $x_i,x_2,\ldots,x_n \in Z_q$ and compute $y_1=g^{x_1},y_2=g^{x_2},\ldots,y_n=g^{x_n}$ and invoke $A$ with $y$ and the additional $y_i$'s. The adversary $A$ can break the binding property means with a non-negligible probability, $A$ can find two different pairs $(m,r)$ and $(m',r')$ such that $g^my^r=g^{m'}y^{r'}$. If $A$ returns such a pair, it must hold that $m+xr=m'+xr'$, and $B$ can compute $x=\frac{m'-m}{r-r'}$, thus solve the DL problem. Since we assume the DL problem is hard, such $A$ does not exist. Therefore the binding property also holds (as Squeamish Ossifrage also said in the comment).