Suppose ${\bf p} = [p_1, \dots, p_n]^T$ be a $\mathbb{C}^{n \times 1}$ vector whose elements are i.i.d zero-mean and unit variance random variable (RVs), i.e., $\mathbb{E}[|p_i|^2] = 1$. Then from law of large numbers we have $\frac{1}{n} {\bf p}^H {\bf p} \to 1$.
Now let ${\bf D} \in \mathbb{R}^{n \times n}$ be a diagonal matrix with different real entries, i.e., $d_{n,n}$. Then can I use law of large numbers to find convergence for:
\begin{equation} \frac{1}{n} {\bf p}^H {\bf D}{\bf p} \end{equation}
My initial calculations is as follows:
Let $X_n = {\bf p}^H {\bf D}{\bf p}$, then $\mathbb{E}[X_n] = \mathbb{E}[\sum_i p_i^* d_{i,i} p_i] = \sum_i \mathbb{E}[|p_i|^2 d_{i,i}] = \sum_i d_{i,i}$, hence we have $\mathbb{E}[{\bf p}^H {\bf D} {\bf p}] = trace({\bf D})$. So can I say $\frac{1}{n} {\bf p}^H {\bf D}{\bf p} \to trace({\bf D})$ as $n \to \infty$.
You've got a tiny mistake, it should go to $\frac{1}{n}\operatorname{tr} D$ because $$\frac{1}{n} p^HDp = \frac{1}{n}\operatorname{tr} (p^HDp) = \frac{1}{n}\operatorname{tr} (p p^HD)= $$ The term $p p^H \rightarrow I$, so you are left with $\frac{1}{n}\operatorname{tr} D$.