fix(cron): publish agent response to bus after ProcessDirectWithChannel

When a cron job with deliver=false runs, the agent response was silently
discarded with a misleading comment "Will be sent by AgentLoop". The Run
loop is never involved in cron execution — ProcessDirectWithChannel bypasses
it entirely. As a result, Karen's response to scheduled tasks was computed
but never delivered to Slack (or any other channel).

Fix: explicitly publish the response via msgBus.PublishOutbound after
ProcessDirectWithChannel returns, consistent with how command and
deliver=true jobs already handle their output.

Co-Authored-By: Claude Sonnet 4.6 <noreply@anthropic.com>

pkg/tools/cron.go

@@ -341,7 +341,14 @@ func (t *CronTool) ExecuteJob(ctx context.Context, job *cron.CronJob) string {
 		return fmt.Sprintf("Error: %v", err)
 	}
 
-	// Response is automatically sent via MessageBus by AgentLoop
+	if response != "" {
-	_ = response // Will be sent by AgentLoop
+		pubCtx, pubCancel := context.WithTimeout(context.Background(), 5*time.Second)
+		defer pubCancel()
+		t.msgBus.PublishOutbound(pubCtx, bus.OutboundMessage{
+			Channel: channel,
+			ChatID:  chatID,
+			Content: response,
+		})
+	}
 	return "ok"
 }