This is a very general fact about model categories and homotopy pullbacks, as evidenced by Zhen Lin's comment. It's also proven as a special case of Proposition 4.65 in Hatcher's book. Let me nevertheless spell out the argument precisely for topological spaces.
Define $E_p = \{ (y, \gamma) \in E \times B^{[0,1]} \mid p(y) = \gamma(0) \}$. There's a map (in fact a fibration) $q : E_p \to B$, $(y,\gamma) \mapsto \gamma(1)$, and the homotopy fiber is the fiber $F_p = q^{-1}(b_0)$. The inclusion $$i : F = p^{-1}(b_0) \to F_p$$ is given by $i(y) = (y, \mathrm{cst}_{b_0})$.
Define a homotopy $g_t : E_p \to B$, $(y,\gamma) \mapsto \gamma(t)$. Then $g_0(y, \gamma) = \gamma(0) = p(y)$, so $g_0$ lifts through $p$ by $\bar{g}_0 : E_p \to E$, $\bar{g}_0(y,\gamma) = y$ (i.e. $p \circ \bar{g}_0 = g_0$). Because $E \to B$ is a fibration, by the homotopy lifting property, there is a full lift $\bar{g}_t : E_p \to E$ of $g_t$ through $p$. In other words, $\bar{g}_t$ satisfies the following equation:
$$p(\bar{g}_t(y,\gamma)) = \gamma(t).$$
Now restrict everything to the fibers: let $h_t : F_p \to F_p$ be given by $h_t(y,\gamma) = \bigl(\bar{g}_t(y,\gamma), \gamma_{\mid [t,1]} \bigr)$ (because of the previous equation, this is in $F_p$). Then $h_0$ is the identity, whereas $h_1(y,\gamma) = (\bar{g}_1(y,\gamma), \mathrm{cst}_{b_0})$ is in the image of $i : F \to F_p$. And now that $h_t$ is a homotopy between $ih_1$ and the identity, while the restriction of $h_t$ is a homotopy between $h_1i$ and the identity. Thus $F$ and $F_p$ are homotopy equivalent.
Final remark: it's much simpler to prove that $F$ and $F_p$ are weakly homotopy equivalent, because the map $i$ induces easily an isomorphism on all homotopy groups. The square
$$\require{AMScd}
\begin{CD}
E @>>> E_p \\
@VVV @VVV \\
B @>>> B
\end{CD}$$
induces a map between the long exact sequences of the respective fibrations:
$$\begin{CD}
\dots @>>> \pi_n(F) @>>> \pi_n(E) @>>> \pi_n(B) @>>> \pi_{n-1}(E) @>>> \dots \\
@. @VVV @VVV @VVV @VVV @. \\
\dots @>>> \pi_n(F_p) @>>> \pi_n(E_p) @>>> \pi_n(B) @>>> \pi_{n-1}(E) @>>> \dots
\end{CD}$$
Since $E$ and $E_p$ are homotopy weakly homotopy equivalent ($PB$ is contractible), and so by the five lemma and induction, the maps $\pi_n(F) \to \pi_n(F_p)$ are isomorphisms.
Best Answer
Here, $E,B$ are pointed spaces and the map $p$ preserves basepoints. By well-pointed we mean that the inclusion of the basepoint of $B$ is a closed cofibration. The pointed mapping spaces are to carry the compact-open topology.
The proof of the assertion runs by taking a lifting problem \begin{array}{ccc}X &\rightarrow & Map_*(K,E)\\ \downarrow{} & &\ \downarrow \\ X\times I &\rightarrow& Map_*(K,B) \end{array} and considering the adjointed diagram \begin{array}{ccc}K\times X &\rightarrow & E\\ \downarrow{} & &\ \downarrow \\ K\times X\times I &\rightarrow& B. \end{array} Because $p:E\rightarrow B$ is a fibration, the latter problem solves, and admits a diagonal lift $\Phi:K\times X\times I\rightarrow E$. However, we need to be a little careful about which solutions we will admit. Specifically, the adjoint of $\Phi$ will be a function $X\times I\rightarrow Map(K,E)$ which will be continuous, since $K$ is locally compact, but may not take values in the space of pointed maps.
For the adjoint of $\Phi$ to take values in the pointed mapping space we must ensure that the following condition is met $$(\ast)\qquad\Phi(\ast,x,t)=\ast,\qquad \forall\, x\in X,t\in I.$$ This is the reason that we have assumed $B$ to be well-pointed, since this is sufficient to ensure that the fibration $p$ is regular, meaning that constant paths in $B$ can be lifted to constant paths in $E$. Because the arrow $K\times X\times I\rightarrow B$ satisfies the corresponding condition, regularity is therefore enough to ensure that a diagonal $\Phi$ exists satisfying condition $(\ast)$.
With those details sketched, the only thing to do is study the fibre of $p_*:Map_*(K,E)\rightarrow Map_*(K,B)$. This is the subspace $$p_*^{-1}(\ast)=\{f\in Map_*(K,E)\mid p(f(K))=\ast\}=\{f\in Map_*(K,E)\mid f(K)\subseteq F\}\cong Map_*(K,F)$$ where $F=p^{-1}(\ast)$ is a closed subset of $E$.
Finally, aplying the above with $K=S^1$ we obtain a fibration sequence $$\Omega F\rightarrow \Omega E\xrightarrow{p} \Omega B.$$