Strong deformation retract of singleton $\{x\}$: there exists a continuous $H: X \times I \to X$ s.t. $\forall t \in I: H(x,t) = x$, $\forall y \in X: H(y,0) = y$ and $\forall y \in X: H(y,1) = x$.
Locally path connected at a point $p$: there exists an open neighborhood basis of $p$ consisting of path connected sets.
So the question is where $\{p\}$ a strong deformation retract implies that $X$ is locally path connected at $p$.
I haven't been able to come up with any counterexamples: all spaces I have consider which are contractible are locally path connected at all points that are strong deformation retracts.
I have prove the intermediate lemma, which would likely factor into to any proof of the affirmative: $\{p\}$ is a strong deformation retract implies that for all open neighborhoods $U$ of $p$, there exists an open neighborhood $V \subseteq U$ of $p$ s.t. $\forall y \in V$ there exists a path from $y$ to $p$ lying entirely in $U$.
This can be proved by considering $H^{-1}(U)$, where $H$ is the homotopy described above, noting that $\{p\} \times I \subseteq H^{-1}(U)$, so by the tube lemma, there exists an open $V \subseteq X$ s.t. $V \times I \subseteq H^{-1}(U)$ and $p \in V$. Then the homotopy induces a path from $y \in V$ to $p$ which lies in $U$.
On the other hand I have exhibited a TS $X$ s.t. there exists a point $p$ s.t. for all open neighborhoods $U$ of $p$, there exists an open neighborhood $V \subseteq U$ of $p$ s.t. $\forall y \in V$ there exists a path from $y$ to $p$ lying entirely in $U$, and yet $X$ is not locally path connected at that point. So the lemma alone isn't sufficient. Unfortunately, this space is not a strong deformation retract (I'm fairly certain) to that point $p$, so it is not a counterexample. I can give a construction of the example if anyone wants it.
It is true. Your "intermediate lemma" is a well-known alternative characterization of local path connectivity.
Let us show that the following are equivalent:
$X$ is locally path connected at $p$.
For all neighborhoods $U$ of $p$, there exists a neighborhood $V \subset U$ of $p$ such that for all $y\in V$ there exists a path from $y$ to $p$ lying entirely in $U$. [If this is satisfied, $X$ is called locally $0$-connected at $p$ (short: $LC^0$ at $p$).]
$1. \Rightarrow 2.$: Trivial.
$2. \Rightarrow 1.$: Let $U$ be a neighborhood of $p$ and $C \subset U$ be the path component of $p$ in $U$. Choose a neighborhood $V \subset U$ of $p$ such that for all $y\in V$ there exists a path from $y$ to $p$ lying entirely in $U$. Then clearly all $y \in V$ are in $C$. Thus $C$ is neighborhood of $p$ and by definition it is path connected.
Note that $C$ is not necessarily an open neighborhood of $p$. If you understand "neigborhood" as a synomyn for "open neighborhood", then the proof does no longer work. However, a space $X$ is locally path connected iff it is $LC^0$ (i.e. $LC^0$ at all points). In fact, now $U$ is open which implies that $U$ is also a neighborhood of each $p'$ in the path component $C \subset U$ of $p$ and we see as above that $p'$ has a neighborhood $V \subset U$ which is contained in $C$.