Suppose $C$ is a component of $X$, where $X$ is locally connected. If $x \in C$, then $x$ has a connected neighbourhood $U_x$. So $C \cup U_x$ is connected, as the union of two intersecting (in $x$) connected sets and $C \subseteq C \cup U_x$. So by maximality $U_x \subseteq C$. So $x$ is an interior point of $C$, and as $x \in C$ was arbitrary, $C$ is open. A connected component is always closed (or $\overline{C}$ would be a strictly larger connected subset). So $C$ is clopen.
This means that if $C$ is a component of $x$ in $X$, $C$ is one of the clopen subsets we intersect in computing the pseudocomponent. So the pseudocomponent $P_x$ of $x$ is a subset of $C$. By general theory, which I hope you have covered, the component of $x$ is a subset of the pseudocomponent $P_x$ of $x$ and so we have equality.
If you have Munkres (2nd ed.): Thm 25.4 states:
A space $X$ is locally path-connected iff for every open set $U$ of $X$, each path-component of $U$ is open in $X$.
So if indeed $U$ is open and connected it has (as all spaces) a decomposition into path-components, which are open in $X$ and thus open in $U$ too, and by being a partition, they are also closed (the complement is also a union of open sets) in $U$. So by connectedness there can be only one path-component.
25.5 even says (part 2 of it)
If $X$ is a topological space, and $X$ is locally path-connected, its components and path-components coincide.
Apply this to $X=U$ (which is locally path-connected as an open subspace of a locally path-connected space) and you're done right away.
Note, this is assuming you use the same definition of local path-connectedness as Munkres does (which is non-standard): every neighbourhood $U$ of $x$ contains a path-connected neighbourhood $V$ of $x$.
The definition in e.g. Engelking is:
for every open set $U$ and every $x \in U$ there is an open neighbourhood $V$ of $x$ such that for any $y \in V$ there is a path $p: [0,1] \to U$ connecting $x$ to $y$.
Note that $V$ is not supposed to be itself path-connected, as Munkres does. So the latter has a stronger notion, so maybe this fact only holds for the stronger notion; at least the proof does.
Best Answer
Let $X$ be the space and fix $p \in X$. Let $C$ be the set of all points in $X$ which are path connected to $p$. Now, it is enough to show that $C$ is closed and open in $X$, to show that $C=X$ (given that $p \in C$ via a constant path, so $C$ is non-empty).
To show $C$ is open, let $c \in C$, then we can choose (by local path connectedness) an open subset $U$ containing $c$. For $u \in U$, $u$ is path connected to $c$ which is path connected to $p$, so by joining paths, we have that $u$ is path connected to $p$. In other words, $U \subset C$, so $C$ is open.
Look at the closure of $C$, namely $\bar{C}$. Let $d \in \bar{C}$, and choose, like above, an open path connected subset $V$ containing $d$. Note that $V \cap C \neq \phi$, because $V$ is open! Hence let $e \in V \cap C$, then $d$ is path connected to $e$ which is path connected to $p$ (because $e \in C$)! Hence $d$ is path connected to $p$ and $d \in C$, so $C=\bar{C}$, $C$ is closed.
Hence $C=X$, and $X$ is path connected.
EDIT : In the style of equivalence relations , we can create another proof. Indeed, define an equivalence relation on $X$ given by $p \sim q$ if there is a path connecting $p$ and $q$. It is easy to see that this is an equivalence relation : for reflexivity, use the constant path. For symmetry, use the reverse of a path, and for transitivity, use the concatenation of the two paths.
Finally, note that every equivalence class is open, because if I take $p$, then by local path connectedness, some neighbourhood of $p$ is path connected, but this includes $p$. Therefore, this entire neighbourhood is in the same equivalence class as $p$. The openness follows.
Now, the equivalence classes partition $X$ into disjoint open sets. Since $X$ is connected, this can't happen unless there's only one equivalence class, which is the whole of $X$. We are done!