$p:E\to B$ is fibration then $p_*:map(X,E)\to map(X,B)$ is fibration as well. 
$p:E\to B$ is fibration then for $X$ being compactly generated weakly Hausdorff space  $p_*:map(X,E)\to map(X,B)$ is fibration as well.

We'd like to show that for any $Y$ and continuous $f$ and homotopy $H$ exists lift $\tilde H$ s.t. following diagram is commutative:
$$
\begin{matrix}
Y & \xrightarrow{f} & map(X,E) \\
\left\downarrow{i_0}\vphantom{\int}\right. & \nearrow{\tilde H}\vphantom{\int}& \left\downarrow{p_*}\vphantom{\int}\right.\\
Y\times I&  \xrightarrow{H} & map(X,B)
\end{matrix}
$$
My attempt was to construct $\tilde H$ using fact that $p$ is fibration, namely for every element $y \in Y$ (which can be done because $Y$ is set):
$$
\begin{matrix}
X & \xrightarrow{f(y)} & E \\
\left\downarrow{i_0}\vphantom{\int}\right. & \nearrow{\tilde H(y,\square)}\vphantom{\int}& \left\downarrow{p}\vphantom{\int}\right.\\
X\times I&  \xrightarrow{H(y,\square)} & B
\end{matrix}
$$
where $H(y,\square): (x,t)\mapsto H(y,t)(x)$. Then homotopy given by formula $\tilde H: (y,t) \mapsto \tilde H(y,t)$ gives us demanded lift in first diagram. Is it correct proof? Where did I use (indirectly) assumption about $X$ being CGWH space or one can get rid of it? My suspicion suggest that this assumption could have used if we had consider all those diagrams in $\mathcal{Top}$ category not $\mathcal{Set}$ (I don't know in which category this problem should hold but the assumption suggests that it may be $\mathcal{Top}$). I'm not familiar with CGWH spaces so I'd be glad for stressing where this assumption plays a role.
 A: Expanding on the comment of Adeel, you have to exploit the following easy fact about orthogonal classes of arrows in categories:
Let $F\dashv G$ be two adjoint functors between categories $\mathcal{C}\leftrightarrows\mathcal{D}$; then $Ff\perp g$ in the category $\mathcal D$ (i.e., $Ff$ has the LLP with respect to $g$ in $\cal D$) if and only if $f\perp Gg$ in the category $\cal C$.
Now apply this to $X\times(-)\dashv Map(X,-)$; the exercise is not finished (it remains showing that $X\times i$ is again an acyclic cofibration!) but I think it's the best hint to work conceptually.
A: My attempt to prove categorical remarks of Adeel and tetrapharmakon from the comments was too long for the comment hence I'm posting it as an answer.

(Lemma) Let $F\dashv G$ be two adjoint functors between categories $\mathcal{C}\leftrightarrows\mathcal{D}$; then $Ff\perp g$ in the category $\mathcal D$ (i.e., $Ff$ has the LLP with respect to $g$ in $\cal D$) if and only if $f\perp Gg$ in the category $\cal C$. (where $f:C_1\to C_2$, $g:D_2 \to D_1$).

In other words there exists an arrow $FC_2 \to D_2$ making following diagram commutative:
$$
\begin{matrix}
FC_1 & \xrightarrow{u} & D_2 \\
\left\downarrow{Ff}\vphantom{\int}\right. & & \left\downarrow{g}\vphantom{\int}\right.\\
FC_2&  \xrightarrow{v} & D_1
\end{matrix}
$$
if and only if there exists an arrow $C_2 \to GD_2$ making following diagram commutative:
$$
\begin{matrix}
C_1 & \xrightarrow{u'} & GD_2 \\
\left\downarrow{f}\vphantom{\int}\right. & & \left\downarrow{Gg}\vphantom{\int}\right.\\
C_2&  \xrightarrow{v'} & GD_1
\end{matrix}
$$
But $F\dashv G$ means that following diagram is commutative:
$$
\begin{matrix}
hom_\mathcal{D}(FC_2,D_2) & \xrightarrow{\simeq} & hom_\mathcal{C}(C_2,GD_2) \\
\left\downarrow{hom_\mathcal{D}(Ff,g)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(f,Gg)}\vphantom{\int}\right.\\
hom_\mathcal{D}(FC_1,D_1)&  \xrightarrow{\simeq} & hom_\mathcal{C}(C_1,GD_1)
\end{matrix}
$$
Once again from adjunction (i.e. isomorphism between hom-sets) we know that to every $u' \in hom_\mathcal{D}(C_1,GD_2)$ corresponds uniquely some $u \in hom_\mathcal{D}(FC_1,D_2)$. Similarly let $v \in hom_\mathcal{C}(FC_2, D_1)$ be a morphism corresponding  to $v' \in hom_\mathcal{C}(C_2,GD_1)$. Let's suppose that there exists $\eta:FC_2 \to D_2$ making the very first with diagram commutative for preceding choice of morphisms $u, v$ and $\phi:C_2 \to GD_2$ morphism corresponding to it by adjunction.
$$
\begin{matrix}
hom_\mathcal{D}(FC_2,D_2) & \xrightarrow{\simeq} & hom_\mathcal{C}(C_2,GD_2) \\
\left\downarrow{hom_\mathcal{D}(Ff,id)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(f,id)}\vphantom{\int}\right.\\
hom_\mathcal{D}(FC_1,D_2)&  \xrightarrow{\simeq} & hom_\mathcal{C}(C_1,GD_2) \\
\left\downarrow{hom_\mathcal{D}(id,g)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(id,Gg)}\vphantom{\int}\right.\\
hom_\mathcal{D}(FC_1,D_1)&  \xrightarrow{\simeq} & hom_\mathcal{C}(C_1,GD_1)
\end{matrix}
$$
for setting above we have:
$$
\begin{matrix}
\eta & \xrightarrow{\simeq} & \phi \\
\left\downarrow{}\vphantom{\int}\right. & & \left\downarrow{}\vphantom{\int}\right.\\
\eta \circ Ff = u &  \xrightarrow{\simeq} & \phi \circ f = u' \\
\left\downarrow{}\vphantom{\int}\right. & & \left\downarrow{}\vphantom{\int}\right.\\
g\circ u = g \circ \eta \circ Ff&  \xrightarrow{\simeq} & Gg \circ \phi \circ f = Gg \circ u'
\end{matrix}
$$
Similarly:
$$
\begin{matrix}
hom_\mathcal{D}(FC_2,D_2) & \xrightarrow{\simeq} & hom_\mathcal{C}(C_2,GD_2) \\
\left\downarrow{hom_\mathcal{D}(id,Gg)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(id,g)}\vphantom{\int}\right.\\
hom_\mathcal{D}(FC_2,D_1)&  \xrightarrow{\simeq} & hom_\mathcal{C}(C_2,GD_1) \\
\left\downarrow{hom_\mathcal{D}(Ff,id)}\vphantom{\int}\right. & & \left\downarrow{hom_\mathcal{C}(f,id)}\vphantom{\int}\right.\\
hom_\mathcal{D}(FC_1,D_1)&  \xrightarrow{\simeq} & hom_\mathcal{C}(C_1,GD_1)
\end{matrix}
$$
so:
$$
\begin{matrix}
\eta & \xrightarrow{\simeq} & \phi \\
\left\downarrow{}\vphantom{\int}\right. & & \left\downarrow{}\vphantom{\int}\right.\\
g \circ \eta  = v &  \xrightarrow{\simeq} & Gg \circ \phi = v' \\
\left\downarrow{}\vphantom{\int}\right. & & \left\downarrow{}\vphantom{\int}\right.\\
v \circ f = g \circ \eta \circ Ff&  \xrightarrow{\simeq} & Gg \circ \phi \circ f =  u' \circ f
\end{matrix}
$$
in the same way we can show that the existence of lifting in second diagram implies existence of lifting in the first and hence the thesis of lemma.
A: Solution which I came up with today:
Let's recall the definition of a cocylinder $P(p)$ i.e. the pullback of such diagram:
$$
\begin{matrix}
P(p) & \xrightarrow{p'} & P(B) \\
\left\downarrow{p_0}\vphantom{\int}\right. & & \left\downarrow{p_0}\vphantom{\int}\right.\\
E&  \xrightarrow{p} & B
\end{matrix}
$$
(where $p':(e,\omega)\mapsto p \circ \omega$).
We may apply this for object $P(E)$ and morphisms $p_* :P(E) \to P(B)$, $p_0:P(E)\to E$ obtaining a unique arrow $\bar p :P(E) \to P(p)$. However we know that $p$ is fibration iff $\exists s:P(p) \to P(E)$ such that $ \bar p  \circ s = id_{P(p)}$.
Let's map our diagram against functor $map(X, \square)$ obtaining:
$$
\begin{matrix}
map(X,P(p)) & \xrightarrow{p'_*} & map(X,P(B)) \\
\left\downarrow{p_{0*}}\vphantom{\int}\right. & & \left\downarrow{p_{0*}}\vphantom{\int}\right.\\
map(X,E)&  \xrightarrow{p_*} & map(X,B)
\end{matrix}
$$
There exists a unique arrow $ \bar p_* : P(map(X,E)) \to map(X,P(p))$ since $map(X,\square)$ preserves limits (hence pullbacks as well) (I'm cheating here a little because we know that there exists some unique arrow not necessarily $\bar p_*$ but it is isomorphic to $\bar p_*$ so let's stick to $\bar p_*$) but moreover in our CGWH category: $$map(X,P(B)) = map(X,map(I,B))\cong map(I \times X,B)\cong map(X \times I,B) \cong map(I ,map(X, B)) = P(map(X,B))$$ so $map(X,P(p))$ is cocylinder $P(p_*)$ and now we have only to show that there exists $s' : map(X,P(p)) \to P(map(X,E))$ such that $\bar p _* \circ s' = id_{P(p_*)}$. Surprisingly $s' = map(X,s)$ works like charm since $\bar p_* \circ map(X, s) = map(X,\bar p \circ s) = map(X,id_{P(p)}) = id_{map(X,P(p))} = id_{P(p_*)}$.
