We already have 2 good solutions here, but wanted to answer using the conventional approach.
Let $\mathbb E(k)$ be expected number of runs in a series of $k$ coin tosses. It is important to remember that the coin tosses are memoryless. Therefore, $\mathbb E(k+1)$ depends only on the value of coin toss $k$ and $k+1$. The value of $\mathbb E(k+1)$ will increase by 1 if $k^{th}$ and $k+1^{th}$ tosses are different.
Keeping this in mind, we translate these into equations
$ \mathbb E(k+1) = p(p\mathbb E(k) + (1-p)(1+ \mathbb E(k)) + (1-p)((1-p)\mathbb E(k) + p(1+ \mathbb E(k))) $
Rearranging this a bit gives us
$ \mathbb E(k+1) = 2p(1-p) + \mathbb E(k) $
Our $ \mathbb E(k) $ is of the form $ \mathbb E(k) = Ck +D $ for some constant $C$ and $D$. The result follows after finding these constants from the previous equation and the initial condition.