Say we want to compute the mean value of $X$ given a probability distribution $P(X)$. Denote this quantity $\langle X \rangle_P$. Then, by definition:
\begin{eqnarray}
\langle X \rangle_P &=& \sum_X X P(X)
\end{eqnarray}
Suppose we draw $M$ samples $\hat{X}_j$ ($i=1,2,...M$) from $P$. Then we can compute an estimate of $\langle X \rangle_P$ using an average over the samples...
\begin{eqnarray}
\langle X \rangle_P &=& \frac{1}{M} \sum_{j=1}^M \hat{X}_j
\end{eqnarray}
Easy enough!
Now suppose we have another probability distribution $Q(X)$. We can express $\langle X \rangle_P$ in terms of an average $\langle \cdots \rangle_Q$ over the distribution $Q$ using the following trickery
\begin{eqnarray}
\langle X \rangle_P &=& \sum_X X P(X) \frac{Q(X)}{Q(X)} \\
&=& \sum_X \left[X\frac{P(X)}{Q(X)}\right] Q(X) \\
&=& \langle X \frac{P}{Q}\rangle_Q
\end{eqnarray}
The basic idea of importance sampling takes advantage of this identity by saying that to compute $\langle X \rangle_P$, it may be more convenient to first draw samples from $Q$, and then correct the naive average you would obtain using the samples from $Q$ by using a weighted average with weights $P/Q$. A typical reason to do this would be if $P$ has a tail which is difficult to sample from directly; you can instead sample from $Q$ which has a more uniform distribution, and apply weights to get the average with respect to $P$.
In more concrete terms, suppose we draw samples $\hat{X}_i$ ($i=1,2,...N$) from $Q$. [Very important note: even though I'm using similar notation, $\hat{X}_i$, $i=1,...,N$, for samples drawn from $Q$ that I used for samples drawn from $P$ above, $\hat{X}_j$, $i=1,...,M$, it's crucially important to understand that these two sets of samples are drawn from different distributions have different statistical properties. From here on out in the answer, I will only refer to samples drawn from $Q$, denoted by $\hat{X}_i$] Then we can estimate $\langle X \rangle_P$ as
\begin{eqnarray}
\langle X \rangle_P &=& \frac{1}{N} \sum_{i=1}^N \hat{X}_i \frac{P(\hat{X}_i)}{Q(\hat{X}_i)}
\end{eqnarray}
Now let's turn to your example. The abstract definitions of $P$ and $Q$ I gave are related to the distributions in your question via
\begin{eqnarray}
P &=& \frac{e^{-\beta E_\mu}}{\sum_\mu e^{-\beta E_\mu}} \\
Q &=& p_\mu
\end{eqnarray}
Note: I relabeled the quantity $Q$ in your question to $X$ to avoid conflicting with the distribution $Q(X)$ I am using in my answer.
We can estimate $\langle X \rangle_P$ using importance sampling via
\begin{eqnarray}
\langle X \rangle_P &=& \langle X \frac{P}{Q} \rangle _Q \\
&=& \langle X \cdot \frac{e^{-\beta E_\mu}}{\left[\sum_\mu e^{-\beta E_\mu}\right]}\cdot \frac{1}{p_\mu} \rangle _Q
\end{eqnarray}
If we draw samples $\hat{X}_i$ ($i=1,2,...,N$) from $Q$, then our estimate of $\langle X \rangle_P$ is
\begin{eqnarray}
\langle X \rangle_P &=& \frac{1}{N} \frac{ \sum_{i=1}^N \hat{X}_i e^{-\beta E_{\mu_i}} p_{\mu_i}^{-1} }{\sum_{\mu} e^{-\beta E_{\mu}}}
\end{eqnarray}
This is almost what you want, except we have an overall factor of $1/N$ and the sum in the denominator is over states $\mu$ and not over samples labeled by $i$.
No matter: the trick is that we can also estimate the partition function (the sum in the denominator over $\mu$) using importance sampling. On the one hand, of course $\langle 1 \rangle_P=1$ (the average of $1$ is $1$). On the other hand, we have that
\begin{eqnarray}
\langle 1 \rangle_P &=& \langle \frac{P}{Q}\rangle_Q
\end{eqnarray}
Or, in terms of the samples:
\begin{eqnarray}
\frac{1}{N} \sum_{i=1}^N \frac{P(\hat{X}_i)}{Q(\hat{X}_i)} = 1
\end{eqnarray}
Plugging in the explicit expressions for your distributions:
\begin{eqnarray}
\frac{1}{N} \frac{\sum_{i=1}^N e^{-\beta E_{\mu_i}} p_{\mu_i}^{-1}}{\sum_\mu e^{-\beta E_\mu}} = 1
\end{eqnarray}
or
\begin{eqnarray}
N \sum_\mu e^{-\beta E_\mu} = \sum_{i=1}^N e^{-\beta E_{\mu_i}} p_{\mu_i}^{-1}
\end{eqnarray}
Using this trick, we have our final expression for the estimate of $\langle X \rangle_P$ written entirely in terms of sums over samples:
\begin{eqnarray}
\langle X \rangle_P &=& \frac{ \sum_{i=1}^N \hat{X}_i e^{-\beta E_{\mu_i}} p_{\mu_i}^{-1} }{\sum_{i=1}^N e^{-\beta E_{\mu_i}} p_{\mu_i}^{-1}}
\end{eqnarray}
which is the expression you wanted.
